Why does logistic function use e rather than 2?
Since you are going to minimize later on the log likelihood, there is actually no big difference between $\log 2^x=x * \log2$ and $\log e^x=x$. You see the difference is simply a constant.
Nevertheless one could argue to use $2^x$ instead of $e^x$ und also use $\log_2$ instead of $\log$ when it comes to the optimizing step. In fact it is possible to use $2^x$ and also many other functions, which show some desired properties.
Which are:


$\lim\limits_{x \rightarrow \infty}{f(x)}=1$
$\lim\limits_{x \rightarrow -\infty}{f(x)}=0$
$f(x) = -f(-x) + 1$, (symmetric in $(0, 0.5)$


Here is an example of suitable functions from wikipedia.