Edit: Using asymptotic to and not equal to as it keeps the meaning intact and makes sense of the relation
Would it be correct to imply from Euler's prime equality function
$x is ~ (asymptotic) to \frac {(\log x)^{\pi(x)}}{\prod \log p}$
where $\pi(x)$ is prime counting function?
$(\sum_{i=0}^x 1/i)$ ~ $(1 + \frac {1}{2} + \frac {1}{2^2} + \frac {1}{2^3} + \dots)(1 + \frac {1}{3} + \frac {1}{3^2} + \frac {1}{3^3} + \dots)(1 + \frac {1}{5} + \frac {1}{5^2} + \frac {1}{5^3} + \dots)\dots$
In famous Euler's prime euqality equation
$\sum x^{-1} = \prod (1-p^{-1})^{-1}$
We have x terms of left hand side but can we write x as
$x$ ~ $(\text{no of terms of 2})(\text{no of terms of 3})(\text{no of terms of 5})\dots$
or as $x$ ~ $\prod (\log_p x)$
If we find the no. of terms it contributes to $x$ say for some power of each prime number say $p$ so for some $k$ we can say $p^k \le x$
i.e $k = \log_p x = \frac {\log x}{\log p}$
Therefore we can rewrite
$x$ ~ $\frac {\log x}{\log 2}$$\frac {\log x}{\log 3}$$\frac {\log x}{\log 5}\dots$
Simplifying
$x$ ~ $\frac {(\log x)^{\pi(x)}}{\log 2 \log 3 \log 5 \dots}$
We can then put upper & lower bounds on ${\log 2 \log 3 \log 5 \dots}$, for eg for lower bound we can say each term is less than equal to $\log 2$.
i.e $x$ ~ $\left(\frac {\log x}{\log 2}\right)^{\pi(x)}$
Taking log on both sides
$\log x$ ~ $\pi(x) \log \frac {\log x}{\log 2}$
$π(x)$ ~ $\frac {\log x}{\log (\frac {\log x}{\log 2})}$
Or over simplifying
$\pi(x)$ ~ $\frac {\log x}{\log \log x}$
Edit:
After reading throgh some old proofs of PNT this answer resonates my thought process https://math.stackexchange.com/a/1892114/1295152
It is true for any finite number of primes $p_1,\dots,p_k$ that, when $x$ is sufficiently large in terms of $\{p_1,\dots,p_k\}$, the number of integers up to $x$ made up of powers of those primes alone is asymptotic to $\prod_{j=1}^k \frac{\log x}{\log p_j}$.
However, there is an error term in this approximation, and if we tried to extend this idea to an unbounded number of primes, the error term would become far larger than the main term, making the approximation unreliable. Thus trying to examine all integers/all primes this way doesn't end up working.