Answer: $f(n)=n$ for all $n \in \mathbb{N}$.
Taking $m=n=1$ in $(*)$ yields $1+f(1)!\mid f(1)!+f(1)$ and hence $1+f(1)!\mid f(1)-1$. Since $|f(1)-1|<f(1)!+1$, this implies $f(1)=1$.
For $m=1$ in $(*)$ we have $n!+1 \mid f(n)!+1$, which implies $n!\leqslant f(n)$, i.e. $f(n) \geqslant n$.
On the other hand, taking $(m, n)=(1, p-1)$ for any prime number $p$ and using Wilson's theorem we obtain $p|(p-1)!+1| f(p-1)!+1$, implying $f(p-1)<p$. Therefore

$$f(p-1)=p-1$$

Next, fix a positive integer $m$. For any prime number $p$, setting $n=p-1$ in (*) yields $(p-1)!+f(m)!\mid(p-1)!+f(m!)$, and hence

$$(p-1)!+f(m)!\mid f(m!)-f(m)!\text { for all prime numbers } p$$

This implies $f(m!)=f(m)!$ for all $m \in \mathbb{N}$, so $(*)$ can be rewritten as $n!+f(m)!\mid f(n)!+f(m)!$. This implies

$$n!+f(m)!\mid f(n)!-n!\text { for all } n, m \in \mathbb{N}$$

Fixing $n \in \mathbb{N}$ and taking $m \in \mathbb{N}$ large enough, we conclude that $f(n)! = n!$, i.e. $f(n)=n$, for all $n \in \mathbb{N}$.
One readily checks that the identity function satisfies the conditions of the problem.