This is a generalization of APMO 2015/2. There is no such function. For the sake of brevity, let $P(a_1,a_2,\dots,a_n)$ denote the assertion $f(a_1)f(a_2)\dots f(a_n)=f(a_1^n a_2^n \dots a_n^n).$ Claim: $f(a)f(b)=f(c)f(d)$ for all $a\neq b, c\neq d, ab=cd$. Proof. For any $a,b,c,d\in S$ such that $a\neq b, c\neq d, ab=cd$, let $x=ab+cd$. $P(ab,c,d,x+1,x+2,\dots,x+n-3)$ gives $$f(ab)f(c)f(d)f(x+1)f(x+2)\dots f(x+n-3)=f\left(\left(abcd\cdot \frac{(x+n-3)!)}{x!}\right)^n\right).$$$P(cd,a,b,x+1,x+2,\dots,x+n-3)$ gives $$f(cd)f(a)f(b)f(x+1)f(x+2)\dots f(x+n-3)=f\left(\left(abcd\cdot \frac{(x+n-3)!)}{x!}\right)^n\right).$$Thus, $f(ab)f(c)f(d)=f(cd)f(a)f(b)$, so $f(c)f(d)=f(a)f(b)$ for all $n\geq 3$. For $n=2$, we can simply compare $P(a,b),P(c,d)$. From our claim, we have $\frac{f(2^m)}{f(2^n)}=\left(\frac{f(4)}{f(2)}\right)^{m-n}=k^{m-n}$ for all $m,n\geq 3$ by simple induction. $P(2^3,2^4,\dots,2^{n+2})$ gives $$f(2^3)f(2^4)\dots f(2^{n+2})=f(2^{\frac{n^3+5n^2}{2}}).$$$P(2^4,2^5,\dots,2^{n+3})$ gives $$f(2^4)f(2^5)\dots f(2^{n+3})=f(2^{\frac{n^3+7n^2}{2}}).$$Hence, $\frac{f(2^{n+3})}{f(2^3)}=\frac{f(2^{\frac{n^3+7n^2}{2}})}{f(2^{\frac{n^3+5n^2}{2}})}$. Using the result from induction, we have $$k^n=k^{n^2}\implies k=1, k=-1\text{ or }n^2-n=0$$Note that $k=\frac{f(4)}{f(2)}$ must be positive. Also, if $n^2-n=0$ then $n=0$ or $1$. Thus, $k=1$. This implies that $f(2^m)=f(2^n)$ for all $m,n\geq 3$. $P(2^3,2^4,\dots, 2^{n+2})$ gives $$f(2^3)f(2^4)\dots f(2^{n+2})=f(2^{\frac{n^3+5n^2}{2}})$$which means that $f(2^3)^n=f(2^3)$ and so $f(2^3)=1\notin S$, a contradiction.