It does not have to. You know $e^x= \sum_0^\infty x^n/ n!$.

Now (using the convention that summing from or to a non-integer number means summing to or from their floor) $$\sum_0^{\sqrt{x}-1} x^n/ n! < \sum_0^{\sqrt{x}-1} x^n< x^{\sqrt{x}}$$ for large $x$,

and

$$\sum_{x^2}^{\infty} x^n/ n! < \sum_{x^2}^{\infty} x^n/ (x^2/e)^n< \sum_{x^2}^{\infty} (e/x)^n < 1$$ for large $x$.

So if we take the subsequence of $1,2,...$ consisting of those integers between all $2^{4k+2}$ and $2^{4k+4}$, but not the rest (that is from 5 to 16, then all from 65 to 256 etc) , then for $x=2^{4k+1}$ (like 2, 32 etc) we will get $\sum_0^\infty x^{a_k}/(a_k)! <x^{\sqrt{x}}+1$, so that $f(x)^{1/x}$ limits to $1$, and for $x=2^{4k+3}$ (like 8, 128 etc) we will get
$\sum_0^\infty x^{a_k}/(a_k)! >e^x -x^{\sqrt{x}}-1$ so that $f(x)^{1/x}$ limits to $e$.

Edit: How to see the inequalities for $x=2^{2k+1}$ and $x=2^{2k+3}$, also known as "what is going on?"

The idea is that for each $x$ the sum $e^x=\sum x^n/n!$ is divided into 3 parts - $n$ from $1$ to $\sqrt{x}$, then $n$ from $\sqrt{x}$ to $x^2$ and then from $x^2$ to $\infty$, and the first and last parts are relatively small.

Now $\sum_0^\infty x^{a_k}/(a_k)!$ is the sum of only those $n$ that $a_k$ picks out form $\mathbb{N}$. So if we make a sequence that does not pick out anything from the middle part for some $x$, that is avoids the $n$ from $\sqrt{x}$ to $x^2$, then the sum for that $x$ would be smaller than the two remaining parts, that is $x^{\sqrt{x}}+1$.

Similarly, if the first and the last parts of the summ are small, then the middle part is big, at least $e^x-x^{\sqrt{x}}-1$. So if for some $x$ the subsequence hits all the integers in the middle part, then the sum will be at least that big for that $x$.

The sequence picking out the integers between $2^{4k+2}$ and $2^{4k+4}$ avoids the middle part for $x=2^{4k+1}$ and hits all of the middle part for $2^{4k+3}$. Hence the bounds, providing two subsequences where limit is 1 and $e$ correspondingly.