I've been thinking about this problem for a bit:
$$\lim_{n \to \infty} \frac{1}{n} \int_1^n \log x \left( \left\lfloor \frac{n}{x-1} \right\rfloor- \sum_{k=1}^\infty \left\lfloor \frac{n}{x^k} \right\rfloor\right) \, dx.$$
If we assume we can apply the Lebesgue dominated convergence theorem, this should tend to a 0, but I haven't found an appropriate dominating function (not even a useful one for GDCT). Instead, I was able to show it is between 0 and 1. Anyone have any thoughts?
$\textbf{Edit:}$ I figured it would be best to explain the motivation. On Wikipedia for "Euler-Mascheroni constant", it provides the identity $$\sum_{p \leq n} \frac{\log p}{p-1} = \log n - \gamma + o(1),$$ without citing a source. Because of this, I took it upon myself to provide my own proof (once in a while, I'll try to look and see if I could find it). Recalling Legendre's formula for $n!$
$$\log n! = \sum_{p \leq n} \log p \sum_{k=1}^\infty \left \lfloor \frac{n}{p^k} \right \rfloor,$$ after a series of manipulations, we end up with the expression
$$\sum_{p \leq n} \frac{\log p}{p-1} - \log n = \frac{1}{n} \log \frac{n!}{n^n} + \frac{1}{n} \sum_{p \leq n} \log p \sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}.$$
An acquaintance of mine and I were able to show $\displaystyle \frac{1}{n} \sum_{p \leq n} \log p \left\{ \frac{n}{p-1} \right\} \to 1-\gamma$; so then we consider the expression
\begin{align*} & \sum_{p \leq n} \log p \sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}\\ &= \sum_{p \leq n} \log p \left\{ \frac{n}{p-1} \right\} + \sum_{p \leq n} \log p \left(\sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}- \left\{ \frac{n}{p-1} \right\}\right) \end{align*}
If we assume $\displaystyle \sum_{p \leq n} \log p \left(\sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}- \left\{ \frac{n}{p-1} \right\}\right) = o(n)$, then we have $$\sum_{p \leq n} \frac{\log p}{p-1} - \log n \to -1 + 1 -\gamma = - \gamma,$$ which is the result we want. Thus, the task becomes proving
$$\lim_{n \to \infty} \frac{1}{n} \sum_{p \leq n} \log p \left(\sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}- \left\{ \frac{n}{p-1} \right\}\right) = 0.$$ Since $\displaystyle \sum_{k=1}^\infty \left\{ \frac{n}{p^k} \right\}- \left\{ \frac{n}{p-1} \right\} = \left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor$, one way of bounding the sum is \begin{align*} 0&\leq \frac{1}{n}\sum_{p \leq n} \log p \left(\sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}- \left\{ \frac{n}{p-1} \right\}\right) \\ &= \frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right) \\ &\leq \frac{1}{n} \int_1^n \log x \left( \left\lfloor \frac{n}{x-1} \right\rfloor- \sum_{k=1}^\infty \left\lfloor \frac{n}{x^k} \right\rfloor\right) \, dx, \end{align*} which explains why we are considering the integral at hand. I have tried another bound that showed the limit lies in the interval $[0,1]$ but, for our purposes, this is insufficient.
Before I proceed, let me preface and reiterate that the original motivation was to show $$\lim_{n \to \infty} \frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right) = 0,$$ and one direction that was considered was the evaluation of the limit of the integral in the original post $$\lim_{n\to \infty} \frac{1}{n}\int_1^n \log(x)\left(\left\lfloor \frac{n}{x-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{x^k} \right\rfloor\right) \, dx.$$ After going back to drawing board, this direction is no longer considered since another proof was provided through different means. That said, we will now prove our original problem.
Using the fact $\lfloor x \rfloor = x - \{ x\}$, observe we have
\begin{align*}\frac{1}{n}\sum_{p \leq n} \log p \left(\sum_{k=1}^\infty \left \{ \frac{n}{p^k}\right\}- \left\{ \frac{n}{p-1} \right\}\right) = \frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right).\end{align*}
Consider $$\sum_{p \leq n} \log p \left\lfloor \frac{n}{p-1} \right\rfloor.$$ Using the definition for the first Chebyshev function $$\vartheta(x) = \sum_{p \leq x} \log p,$$ we find that \begin{align*} \sum_{p \leq n} \log p\left\lfloor \frac{n}{p-1} \right\rfloor= \sum_{p \leq n+1} \log p \left\lfloor \frac{n}{p-1} \right\rfloor - 1_\mathbb{P}(n+1)\log(n+1), \end{align*} so that we have \begin{align*} \sum_{p \leq n+1} \log p \left\lfloor \frac{n}{p-1} \right\rfloor & = \sum_{p -1 \leq n} \log p \left\lfloor \frac{n}{p-1} \right\rfloor \\ &= \sum_{i = 1}^\infty \sum_{\frac{n}{i+1} < p -1 \leq \frac{n}{i}} \log p \left\lfloor \frac{n}{p-1} \right\rfloor \\ &= \sum_{i = 1}^\infty i \sum_{\frac{n}{i+1} < p -1 \leq \frac{n}{i}} \log p \\ &= \sum_{i = 1}^\infty i \left( \vartheta\left(1+\frac{n}{i}\right) - \vartheta\left(1+\frac{n}{i+1}\right)\right). \end{align*} We apply a similar procedure to the sum $$\sum_{p \leq n} \log p \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor,$$ which, in this instance, we will be using the following relationship between the second Chebyshev function $\psi(x)$ and the first Chebyshev function: $$\psi(x) = \sum_{k=1}^\infty \vartheta(x^{1/k}).$$ Before we move further, we would like to observe that $$\sum_{p \leq n} \log p \left\lfloor \frac{n}{p^k} \right\rfloor - \sum_{p^k \leq n} \log p \left\lfloor \frac{n}{p^k} \right\rfloor = \sum_{n^{1/k} < p \leq n} \log p \left\lfloor \frac{n}{p^k} \right\rfloor = 0.$$ Thus, we have the following \begin{align*} \sum_{p \leq n} \log p \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor &= \sum_{k=1}^\infty \sum_{p \leq n} \log p \left\lfloor \frac{n}{p^k} \right\rfloor \\ &=\sum_{k=1}^\infty \sum_{p^k \leq n} \log p \left\lfloor \frac{n}{p^k} \right\rfloor \\ &=\sum_{i=1}^\infty \sum_{k=1}^\infty \sum_{\frac{n}{i+1} < p^k \leq \frac{n}{i}} \log p \left\lfloor \frac{n}{p^k} \right\rfloor \\ &= \sum_{i=1}^\infty i \sum_{k=1}^\infty \sum_{\frac{n}{i+1} < p^k \leq \frac{n}{i}} \log p \\ &= \sum_{i=1}^\infty i \sum_{k=1}^\infty \vartheta\left(\sqrt[k]{\frac{n}{i}}\right) - \vartheta\left(\sqrt[k]{\frac{n}{i+1}}\right) \\ &= \sum_{i=1}^\infty i \left(\psi\left(\frac{n}{i}\right) - \psi\left(\frac{n}{i+1}\right) \right) \end{align*} Thus, our original sum equals \begin{align*}\frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right) &= -\frac{1_\mathbb{P}(n+1)\log(n+1)}{n} \\ &+\frac{1}{n}\sum_{i = 1}^\infty i \left( \vartheta\left(1+\frac{n}{i}\right) - \vartheta\left(1+\frac{n}{i+1}\right) - \psi\left(\frac{n}{i}\right) + \psi\left(\frac{n}{i+1}\right) \right)\end{align*} We can now turn our attention to the infinite sum; we can rewrite this as a telescoping sum \begin{align*} &\sum_{i = 1}^\infty i \left( \vartheta\left(1+\frac{n}{i}\right) - \vartheta\left(1+\frac{n}{i+1}\right) - \psi\left(\frac{n}{i}\right) + \psi\left(\frac{n}{i+1}\right) \right) \\ &= \sum_{i = 1}^\infty i\vartheta\left(1+\frac{n}{i}\right) - (i+1)\vartheta\left(1+\frac{n}{i+1}\right) - i\psi\left(\frac{n}{i}\right) + (i+1)\psi\left(\frac{n}{i+1}\right)\\ &+ \sum_{i = 1}^\infty \vartheta\left(1+\frac{n}{i+1}\right) - \psi\left(\frac{n}{i+1}\right) \\ &= \vartheta(n+1) - \psi(n) + \lim_{m \to \infty} \left(m\psi\left(\frac{n}{m}\right) - m\vartheta\left(1+\frac{n}{m}\right)\right) \\ &+ \sum_{i = 1}^\infty \vartheta\left(1+\frac{n}{i+1}\right) - \psi\left(\frac{n}{i+1}\right) \\ &= \lim_{m \to \infty} \left(m\psi\left(\frac{n}{m}\right) - m\vartheta\left(1+\frac{n}{m}\right)\right) + \sum_{i = 1}^\infty \vartheta\left(1+\frac{n}{i}\right) - \psi\left(\frac{n}{i}\right) \\ \end{align*} Since $\psi\left(\frac{n}{m}\right)$ and $\vartheta\left(1+\frac{n}{m}\right)$ are both $0$ for $m > n$ for fixed $n$, the limit $$\lim_{m \to \infty} \left(m\psi\left(\frac{n}{m}\right) - m\vartheta\left(1+\frac{n}{m}\right) \right) = 0.$$ Similarly, we have $$\sum_{i = 1}^\infty \vartheta\left(1+\frac{n}{i}\right) - \psi\left(\frac{n}{i}\right) = \sum_{i = 1}^n \vartheta\left(1+\frac{n}{i}\right) - \psi\left(\frac{n}{i}\right).$$ Altogether, we have $$\frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right) = -\frac{1_\mathbb{P}(n+1)\log(n+1)}{n} + \frac{1}{n}\sum_{i = 1}^n \vartheta\left(1+\frac{n}{i}\right) - \psi\left(\frac{n}{i}\right).$$ Taking limits as $n \to \infty$, notice the right hand-sum is a Riemann sum; we have $$\lim_{n \to \infty} \frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right) = \int_0^1 \vartheta\left(1+\frac{1}{x}\right) - \psi\left(\frac{1}{x}\right) \, dx.$$ We now move our efforts to the evaluation of the integral; $u$-substituting $x = 1/u$, $dx = -1/u^2 du$, we have $$\int_0^1 \vartheta\left(1+\frac{1}{x}\right) - \psi\left(\frac{1}{x}\right) \, dx = \int_1^\infty \frac{\psi(x) - \vartheta(x+1)}{x^2} \, dx.$$ Rewrite this integral as $$\int_1^\infty \frac{\psi(x) - \vartheta(x+1)}{x^2} \, dx = \int_1^\infty \frac{\psi(x) - \vartheta(x)}{x^2} \, dx - \int_1^\infty \frac{\vartheta(x+1)-\vartheta(x)}{x^2}\, dx$$ Neither of these integrals diverge so the use of linearity of integrals in this instance is justified; this comes from the fact that $\vartheta(x+1)-\vartheta(x) \leq \log(x+1)$ and from Theorem 4.1 in Tom Apostol's "Introduction to Analytic Number Theory":
Recalling earlier the relationship between the first and second Chebyshev function, the integral on the left is \begin{align*} \int_1^\infty \frac{\psi(x) - \vartheta(x)}{x^2} \, dx &= \int_1^\infty \sum_{k=2}^\infty \frac{\vartheta(\sqrt[k]{x})}{x^2} \, dx \\ &= \int_1^\infty \sum_{k=2}^\infty \frac{\vartheta(u)}{u^{2k}} \, ku^{k-1}du , \quad \quad x = u^k \\ &= \int_1^\infty \frac{\vartheta(u)}{u}\sum_{k=2}^\infty \frac{k}{u^{k}} \, du \\ &= \int_1^\infty \frac{\vartheta(u)}{(u-1)^2} - \frac{\vartheta(u)}{u^2} \, du \end{align*} Back to our integral, we have \begin{align*} \int_1^\infty \frac{\psi(x) - \vartheta(x)}{x^2} \, dx - \int_1^\infty \frac{\vartheta(x+1)-\vartheta(x)}{x^2}\, dx &= \int_1^\infty \frac{\vartheta(x)}{(x-1)^2} - \frac{\vartheta(x)}{x^2} \, dx - \int_1^\infty \frac{\vartheta(x+1)-\vartheta(x)}{x^2}\, dx \\ &= \int_1^\infty \frac{\vartheta(x)}{(x-1)^2} - \frac{\vartheta(x+1)}{x^2}\, dx \end{align*} Before we proceed, we consider the integral $$\int_1^t \frac{\vartheta(x)}{(x-1)^2} - \frac{\vartheta(x+1)}{x^2}\, dx$$ for real $t > 1$. We can rewrite this \begin{align*} \int_1^t \frac{\vartheta(x)}{(x-1)^2} - \frac{\vartheta(x+1)}{x^2}\, dx &= \int_1^t \frac{\vartheta(x)}{(x-1)^2}\, dx - \int_1^t\frac{\vartheta(x+1)}{x^2}\, dx\\ &= \int_1^t \frac{\vartheta(x)}{(x-1)^2}\, dx - \int_2^{t+1}\frac{\vartheta(x)}{(x-1)^2}\, dx\\ &= \int_1^2 \frac{\vartheta(x)}{(x-1)^2}\, dx - \int_t^{t+1}\frac{\vartheta(x)}{(x-1)^2}\, dx\\ &= \int_1^2 \frac{\vartheta(x)}{(x-1)^2}\, dx - \int_0^{1}\frac{\vartheta(x+t)}{(x+t-1)^2}\, dx. \end{align*} Since $\vartheta(x) = 0$ for $x < 2$, we have \begin{align*} \int_1^t \frac{\vartheta(x)}{(x-1)^2} - \frac{\vartheta(x+1)}{x^2}\, dx &= - \int_0^{1}\frac{\vartheta(x+t)}{(x+t-1)^2}\, dx. \end{align*} Now, in "Estimates of some functions over primes without R.H", Pierre Dusart has shown that $\vartheta(x) < Cx$ for some $C>1$. Using this, we find \begin{align*} \int_0^{1}\frac{\vartheta(x+t)}{(x+t-1)^2}\, dx &\leq C\int_0^{1}\frac{x+t}{(x+t-1)^2}\, dx \\ &= C\left(\frac{1}{t(t-1)}-\log\left(1-\frac{1}{t}\right)\right). \end{align*} Thus, by the squeeze theorem, we find \begin{align*} 0 &\geq \lim_{t \to \infty} - \int_0^{1}\frac{\vartheta(x+t)}{(x+t-1)^2}\, dx \geq \lim_{t \to \infty} C\left(\log\left(1-\frac{1}{t}\right)-\frac{1}{t(t-1)}\right) = 0, \end{align*} so our original integral is $$\int_1^\infty \frac{\vartheta(x)}{(x-1)^2} - \frac{\vartheta(x+1)}{x^2}\, dx = 0.$$
Thus, we have $$\lim_{n \to \infty} \frac{1}{n}\sum_{p \leq n} \log p \left(\left\lfloor \frac{n}{p-1} \right\rfloor - \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor\right) = 0,$$ as desired.