$$(2n)! = \binom{2n}{n} \cdot n!^2 \equiv \binom{2n}{n} \cdot (-a)^2 \mod (n! + a),$$
so if $n! + a$ divides $(2n)!$, then it also divides $a^2 \binom{2n}{n}$. We will show that when $n$ is large, $n! + a$ is greater than $a^2 \binom{2n}{n}$ and therefore does not divide it (unless $a = 0$). Assume in the following that $a \neq 0$ and $n > \frac{4^5}{12}a^2$. First of all, by the binomial theorem,
$$\binom{2n}{n} \le \sum_{k=0}^{2n} \binom{2n}{k} = 2^{2n} = 4^n.$$
If $a > 0$, we have
$$0 < \frac{a^2}{n! + a} \binom{2n}{n} < \frac{a^2}{n!} \binom{2n}{n} \le \frac{a^2}{n!} 4^n = \left( \frac{4^{n-5}}{5 \cdot 6 \cdots (n-1)} \right) \frac{4^5 a^2}{24n} < 1.$$
If $a < 0$, then we observe that $n! > n > 2|a|$, so that
$$0 < \frac{a^2}{n! + a} \binom{2n}{n} = \frac{a^2}{n! - |a|} \binom{2n}{n} < \frac{a^2}{\frac{1}{2}n!} \binom{2n}{n} \le \frac{2a^2}{n!} 4^n = \left( \frac{4^{n-5}}{5 \cdot 6 \cdots (n-1)} \right) \frac{4^5 a^2}{12n} < 1.$$

Thus $\frac{a^2 \binom{2n}{n}}{n!+a}$ is not an integer for all such $n$, which contradicts our assumption that $n! + a$ divides $(2n)!$ (and thus $a^2 \binom{2n}{n}$) for infinitely many $n$.
So $a = 0$ is the only possibility (and indeed $n!$ divides $(2n)!$ for all $n$).

We show that $n! + a$ cannot divide $(2n)!$ if $a \neq 0$, $n \ge 2|a|$ and $n \ge 9$. In this case, $\frac{n!}{a}$ is an integer. If $n! + a$ divides $(2n)!$, then so does $\frac{n!}{a} + 1$. Next note that every prime $p \le n$ divides $\frac{n!}{a}$: if $p \nmid a$, this is obvious (since $p \nmid n!$), and if $p \mid a$, then we see that $p$ divides $2|a|$, which occurs as a factor of $\frac{n!}{a}$. In either case, we find that $\frac{n!}{a} + 1$ is not divisible by $p$. So $\frac{n!}{a} + 1$ does not have any prime factors $\le n$, which means that it is coprime to all numbers $\le n$ as well as all the even numbers $\le 2n$.
Therefore $\frac{n!}{a} + 1$ must divide the product of all odd numbers greater than $n$ and less than $2n$. By our assumption, $\left|\frac{n!}{a} + 1\right| \ge \frac{n}{a} \cdot (n-1)! - 1 \ge 2(n-1)! - 1$. We show that the product of the odd numbers between $n$ and $2n$ is less than $2(n-1)! - 1$ for $n \ge 9$, which yields the desired contradiction.
For $n = 9$ and $n = 10$, we have the two inequalities
$$2 \cdot 8! - 1 = 80639 > 36465 = 11 \cdot 13 \cdot 15 \cdot 17$$
and
$$2 \cdot 9! - 1 = 725759 > 692835 = 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$$
respectively. Now we proceed by induction. Assume that
$$2(n-1)! - 1 > \prod_{\substack{n<k<2n \\ k \text{ odd}}} k.$$
If $n > 10$ is even, we use the induction hypothesis to obtain
$$\prod_{\substack{n<k<2n \\ k \text{ odd}}} k = \frac{(2n-3)(2n-1)}{n-1} \prod_{\substack{n-2<k<2n-4 \\ k \text{ odd}}} k < 2(2n-1) \prod_{\substack{n-2<k<2n-4 \\ k \text{ odd}}} k < (4n-2)(2(n-3)!-1) \\ < (n-1)(n-2)(2(n-3)!-1) = 2(n-1)! - (n-1)(n-2) < 2(n-1)!-1,$$
and if $n > 10$ is odd, we get analogously
$$\prod_{\substack{n<k<2n \\ k \text{ odd}}} k = \frac{(2n-3)(2n-1)}{n} \prod_{\substack{n-2<k<2n-4 \\ k \text{ odd}}} k < 2(2n-3) \prod_{\substack{n-2<k<2n-4 \\ k \text{ odd}}} k < (4n-6)(2(n-3)!-1) \\ < (n-1)(n-2)(2(n-3)!-1) = 2(n-1)! - (n-1)(n-2) < 2(n-1)!-1.$$

This completes the induction and thus the proof.