Let $\pi(x)$ be the number of primes not exceeding $x$ and $Li(x) = pv (\int_{1}^{x} \frac{dt}{\log t})$. Consider the prime zeta function.
(Trivial) Claim: For $\Re(s)=\sigma>1/2$, one has
$$s\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log((s-1)\zeta(s))=\sum_{m=2}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$$.
Proof: Consider the prime zeta function $$\sum_{p} p^{-s} = \sum_{m=1}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$$ for $\Re(s)=\sigma>1$, where $\mu$ and $\zeta$ denote the Mobius and Riemann zeta functions, respectively. Li(x) is the Cauchy principal value for $\int_{1}^{x} \frac{dt}{\log t}$ for $x>1$.
Applying partial summation to the left-hand side sum over primes $p$ together with the identity $\int_{1}^{\infty} s Li(x)x^{-s-1} \mathrm{d}x=-\log(s-1)$ for $\sigma>1$ yields
$$s\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x-\log((s-1)\zeta(s))=\sum_{m=2}^{\infty} \frac{\mu(m)}{m}\log \zeta(ms)$$ for $\sigma>1$. Let $\Pi(x)=\pi(x)+\frac{1}{2}\pi(x^{1/2})+\frac{1}{3}\pi(x^{1/3}) + \cdots$. Since $\log \zeta(s)=s \int_{1}^{\infty} \Pi(x)x^{-s-1} \mathrm{d}x$ for $\sigma>1$, it follows that hence $\log((s-1)\zeta(s))=s \int_{1}^{\infty} (\Pi(x)-Li(x))x^{-s-1} \mathrm{d}x$. Thus the LHS of the preceding equation is identical to $s\int_{1}^{\infty} (\pi(x) - \Pi(x))x^{-s-1} \mathrm{d}x$. Note that $|\pi(x)-\Pi(x)| \ll x^{1/2}$. For $\sigma>1/2$ and all $m\geq 2$, we also have $\zeta(ms)=1+2^{-ms}+3^{-ms}+...$ thus $|\mu(m)\log \zeta(ms) | \ll 2^{-m \sigma}$ for $\sigma>1/2$ and all $m\geq 2$. These bounds imply that both sides of the preceding equation are analytic everywhere in the larger plane $\sigma>1/2$. Thus the desired result follows (?) by the identity theorem for holomorhic functions ?
Remark: The above claim is being said to be trivial, in the sense that it doesn't seem to lead to any significant result. For one, it certainly doesn't imply the RH, at lest on its own, since the convergence of the RHS of the above identity for $\Re(s)>1/2$ doesn't rule out the possibility of some zero to the right of the critical line.
$$\int_1^\infty x^\rho x^{-s-1}dx-\frac1{s-\rho}$$ converges only for $\Re(s)> \rho$ and in there it vanishes. This doesn't tell us anything about $\rho$. This is exactly the same situation as in your questions