The proposals are $p_n\sim n\log n$ and $\pi(x)\sim \frac{x}{\log x}$. Consider:
$$p_{\pi(x)}\sim \pi(x)\log\pi(x)\sim\frac{x}{\log x}\log\left(\frac{x}{\log x}\right)=\frac{x}{\log x}\left(\log x-\log\log x\right)\sim x$$
and
$$\pi(p_n)\sim\pi(n\log n)\sim\frac{n\log n}{\log(n\log n)}=\frac{n\log n}{\log n+\log\log n}\sim n$$
Try formailizing this (giving more meaning to $\sim$) to show that $p_{({.})}$ and $\pi({.})$ are asymptotic inverses.
To formalize more: Let $f(x)=\frac{x}{\log{x}}$ and $g(x)=x\log{x}$. Then
$$\frac{f(g(x))}{x}=\frac{1}{x}\frac{x\log{x}}{\log(x\log{x})}=\frac{1}{1+\frac{\log\log x}{\log x}}\xrightarrow{x\to\infty}1$$
And $$\frac{g(f(x))}{x}=\frac{1}{x}\frac{x}{\log{x}}\log\left(\frac{x}{\log{x}}\right)=\frac{1-\frac{\log\log x}{\log x}}{1}\xrightarrow{x\to\infty}1$$
And so for large $x$, $f(g(x))\sim x$ and $g(f(x))\sim x$, where $A\sim B$ means that their ratio approaches $1$.
All this is to say that $f$ and $g$ are asymptotic inverses. So if we start with the premise that the $n$th prime number $p_n$ is approximately $g(n)$ (in the same "ratio approaches 1" sense), and then define $\pi(x)$ as the inverse concept (the number of primes less than a given $x$), then $\pi(x)$ is approximately $f(x)$ (again in that sense).
Best Answer
The number $e^{\gamma}$ plays an important role in the theory of $y$-rough numbers , where the Buchstab-function can be used to estimate the number of $y$-rough numbers below $x$. And this function tends quickly to $e^{-\gamma}$
It is also useful to estimate the probability that a large random number $x$ is prime if it has no prime factor below $y$ , assuming $y$ itself is large , but $x$ much larger.
Neither the Euler number $e$ nor the Euler Mascheroni constant $\gamma$ are however important for all this, just the combined number.