We prove that there is no such function. For arbitrary elements $a$ and $b$ of $S$, choose an integer $c$ that is greater than both of them. Since $b c > a$ and $c > b$, we have
$$f\left(a^{4} b^{4} c^{4}\right) = f\left(a^{2}\right) f\left(b^{2} c^{2}\right) = f\left(a^{2}\right) f(b) f(c)$$
Furthermore, since $a c > b$ and $c > a$, we have
$$f\left(a^{4} b^{4} c^{4}\right) = f\left(b^{2}\right) f\left(a^{2} c^{2}\right) = f\left(b^{2}\right) f(a) f(c)$$
Comparing these two equations, we find that for all elements $a$ and $b$ of $S$,
$$f\left(a^{2}\right) f(b) = f\left(b^{2}\right) f(a) \quad \Longrightarrow \quad \frac{f\left(a^{2}\right)}{f(a)} = \frac{f\left(b^{2}\right)}{f(b)}$$
It follows that there exists a positive rational number $k$ such that
$$\begin{equation*} f\left(a^{2}\right) = k f(a), \quad \text{ for all } a \in S. \tag{1} \end{equation*}$$
Substituting this into the functional equation yields
$$\begin{equation*} f(a b) = \frac{f(a) f(b)}{k}, \quad \text{ for all } a, b \in S \text{ with } a \neq b. \tag{2} \end{equation*}$$
Now combine the functional equation with equations (1) and (2) to obtain
$$f(a) f\left(a^{2}\right) = f\left(a^{6}\right) = \frac{f(a) f\left(a^{5}\right)}{k} = \frac{f(a) f(a) f\left(a^{4}\right)}{k^{2}} = \frac{f(a) f(a) f\left(a^{2}\right)}{k}, \quad \text{ for all } a \in S.$$
It follows that $f(a) = k$ for all $a \in S$. Substituting $a = 2$ and $b = 3$ into the functional equation yields $k = 1$, however $1 \notin S$ and hence we have no solutions.