Plugging $(a, b) = (1, 1)$ to obtain $f(1) = 1$. We can then inductively prove that $f(n) = n$. Assume that it has already been held for all positive integers less than $n$. Plugging $(a, b) = (n, 1)$, then after a simple algebra it follows that $f(n) \le n$ and $(n-1)!f(n) + 1 \mid (n-1)!(n - f(n))$. Since $\gcd((n-1)!, 1 + f(n)(n-1)) = 1$ it follows that $(n-1)!f(n) + 1 \mid n - f(n)$. Thus, $|n - f(n)| \ge (n-1)!f(n) + 1$. Assume on the contrary that $f(n) \ne n$ then $|n - f(n)| \le n - 1 < (n-1)!(n-1) < 1 + (n-1)!f(n)$, a contradiction. Thus, $f(n) = n$ and we are done.