Prime Number Theorem estimate Update
I have updated this question in light of the illuminating answers given below, which clearly point out my mistake. However, I still maintain that Legendre's original guess had some validity - it would seem, as the asymptotic starting point for the plot $\log x - \frac{x}{\pi(x)}$ as given below:

It clearly will begin to converge to $1$ eventually (_Mathematica_ has a limit of not much more than PrimePi[$1 \times 10^{14}$]), but it does appear to have a starting point of around Legendre's first guess, so my initial question still holds in this regard.
As an additional question, does anyone know how far along the numberline one would have to go for $\log x - \frac{x}{\pi(x)}$ to reach close to (let's say $\pm 0.01$) $1$? Is there an asymptotic statement for this?
Original question
Could it be that the bounds for Stirling's approximation, given as $\frac{e}{\sqrt{2\pi}} = 1.0844375514...$ provide a more accurate 'guess' than Legendre's constant of $1.08366$ as an estimate for the convergence of the prime counting function as illustrated here?
The discrepancy is illustrated below in a plot of $\log x - \frac{x}{\pi(x)}$:

Compare:

If this were the case, would it not give a closed form to the approximation?
Edit
A plot of $\frac{x}{\pi(x)}$, $\log x+1.084437\dots$ and $\log x+1$ in response to Eric Naslund's answers below.


No, your value is a worse approximation. The true approximation/value of Legendre's constant is $1$.
By the prime number theorem, $$\pi(x)=\frac{x}{\log x}+\frac{x}{\log^{2}x}+O\left(\frac{x}{\log^{3}x}\right)=\frac{x}{\log x}\left(1+\frac{1}{\log x}+O\left(\frac{1}{\log^{2}x}\right)\right)$$ and so $$\frac{x}{\pi(x)}=\log x\left(1+\frac{1}{\log x}+O\left(\frac{1}{\log^{2}x}\right)\right)^{-1} $$
$$=\log x-1+O\left(\frac{1}{\log x}\right).$$ Your choice of $B$  approximates $\frac{x}{\pi(x)}$  by $\log x-1.084437\dots,$ and since $1.08366$  is closer to $1$, the true value, than $1.084437$, your constant will be a worse asymptotically.
