If $n$ is a prime power $q$ then we can use the polynomials of degree at most $d \rightarrow \infty$ on a finite field $k$ of $q$ elements, because for each $k \leq d$ the expected number of $d$-tuples of fixed points is the same as for a random function. If $n=q+1$ then the same should work for rational functions of degree at most $d$ on the projective line over $k$.

As for alternating permutations, what counts as a simple proof? For each $k \leq n-2$, the expected number of $k$-tuples of fixed points of a random permutation is the same for $A_n$ as for $S_n$, and that's more than enough to prove that the proportion of permutations with exactly $j$ fixed points approaches $1/(j!e)$ for each $j=0,1,2,3,\ldots$. Also, as you know one can get an exact count by computing the generating function for the difference between the number of even and odd derangements.

For $j=0,1,2,\ldots$ let $b_n(j)$ be the number of functions in $C_n$ with exactly $j$ fixed points. At least in some cases when $b_n(0) / a_n \rightarrow 1/e$ (including permutations, functions, and also even permutations and two further examples given below), we have the more general result for each $j$ that $b_n(j) / a_n \rightarrow 1/(j!e)$. That is, if we construct a random variable $j$ by choosing $f \in C_n$ uniformly at random and counting its fixed points, then the distribution of $j$ approaches the Poisson distribution with parameter $1$.

For this distribution, not only does $j$ have expected value $1$, but more generally for each $i$ the expected number of $i$-tuples of distinct fixed points approaches $$ \sum_{j=i}^\infty \frac{j!}{(j-i)!} \frac1{j!e} = \frac1e \sum_{j=i}^\infty \frac1{(j-i)!} = 1. $$ Since the Poisson distribution is determined by its moments [what's a standard reference for this?], it follows that conversely

Suppose for each $n$ in a sequence with $n \rightarrow \infty$ we have a set $C_n$ of functions $f : \lbrace 1,2,\ldots,n \rbrace \rightarrow \lbrace 1,2,\ldots,n \rbrace$. If for each $i$ the average number over $f \in C_n$ of $i$-tuples of distinct points of $f$ approaches $1$ as $n \rightarrow \infty$ then $b_n(j) / a_n \rightarrow 1/(j!e)$ for each $j$.

The hypothesis is reasonably natural in contexts where we expect that the $i$-tuples $\bigl(f(x_1),f(x_2),\ldots f(x_i)\bigr)$ are nearly equidistributed for most $i$-tuples $(x_1,x_2,\ldots,x_i)$ of distinct elements of $\lbrace 1,2,\ldots,n \rbrace$.

This equidistribution holds exactly for all $i \leq n$ if $C_n$ is the set of all functions; if $C_n$ is the permutations then we have exact equidistribution among $i$-tuples of distinct elements of $\lbrace 1,2,\ldots,n \rbrace$, which is sufficient as $n \rightarrow \infty$; likewise for even permutations once $n \geq i+2$, which can be assumed since in each case we fix $i$ and let $n \rightarrow \infty$.

A further example: let $n$ be a prime power $q$, identify $\lbrace 1,2,\ldots,n \rbrace$ with a finite field $F$, and let $C_n$ consist of the $n^d$ polynomials of degree less than $d$. Then equidistribution holds exactly for each $i \leq d$. Once $d>1$, the distribution of $j$ is the same as the distribution of the number of roots of a random polynomial of degree $<d$ (namely $f(X)-X$). It is well known that there is a close link between the distribution of cycle structures of random permutations, and of degrees of irreducible factors of a random polynomial over a finite field (e.g. this is a manifestation of the Čebotarev density theorem); fixed points are 1-cycles, which correspond to linear factors, i.e. roots in $F$.

To conclude, a similar but more exotic example: if $n=q+1$ we can identify $\lbrace 1,2,\ldots,n \rbrace$ with the finite projective line $F \cup \lbrace \infty \rbrace$, fix $d < n/2$, and let $C_n$ be the set of rational functions of given degree $d>1$. Here $a_n = q^{2d+1} - q^{2d-1}$. (See the bottom of page 8 of my STOC'01 paper for a bijective proof that there are exactly $q^{2d+1}$ rational functions of degree at most $d$.) The number of fixed points is at most $d+1$, and for each $i \leq d+1$ the expected number of fixed $i$-tuples behaves correctly. So this example works as long as $d \rightarrow\infty$.