Let $\sum_{p \leq x}$ denote the summation over the primes less than $x$. Let
$$ \theta(x) = \sum_{p \leq x} \ln p. $$
I want to show
$$ \fbox{$\sum_{n \leq x} \frac{\theta(n)}{n^2} = \ln x + O(1).$} $$
I don't think I'm allowed to use the Prime Number Theorem. If I could, the Prime Number Theorem is equivalent to $\theta(n) \sim n$, and I can use the harmonic sum to complete the question.
What I've tried so far is the following: the statement is equivalent to:
$$ \sum_{n \leq x} \theta(n) \frac{x^2}{n^2} = x^2\ln x + O(x^2). $$ If $x$ is a real number, let $[x]$ denote the integer part of $x$, or equivalently, the floor of $x$. I can show that:
$$ \sum_{n \leq x} \theta(n) \frac{x^2}{n^2} = \sum_{n \leq x} \theta(n) \left[\frac{x}{n} \right]^2 + O(x^2). $$ Using the Dirichlet hyperbola method, I can show that $$ \sum_{n \leq x} \theta(n) \left[\frac{x}{n} \right]^2 = \sum_{n \leq x} \theta \star f, $$ where $\star$ denotes Dirichlet convolution and $f$ is the arithmetical function $f(n) = 2n -1$. However, I am unable to proceed because the behavior of $\theta$ is too unpredictable. I've skipped some of the steps because I don't think I'm on the right track, but please do let me know if you would like the full working.
Any hints or insights to this question is appreciated.
This can be proved from Mertens' theorem, which is known to be weaker than the PNT:
$$ \sum_{p\le x}{\log p\over p}=\log x+O(1) $$
See this answer for details.