jmc

Fix $s\in S.$ Setting $y=s-x,$ we get $$f(x)+f(s-x)=f(x(s-x)f(s)).(\ast )$$This holds for all $x\in S,$ $x\ne s.$

Consider the equation $$s-x=x(s-x)f(s).$$The solutions in $x$ are $x=s$ and $x=\frac{1}{f(s)}.$ Since $x\in S,$ $f(s)$ is well-defined. Furthermore, $f(s)\ne 0,$ so $\frac{1}{f(s)}$ is well-defined. If $f(s)\ne \frac{1}{s},$ then we can set $x=\frac{1}{f(s)}$ in $(\ast ),$ which gives us $$f\left(\frac{1}{f(s)}\right)+f\left(s-\frac{1}{f(s)}\right)=f\left(s-\frac{1}{f(s)}\right).$$Then $f\left(\frac{1}{f(s)}\right)=0,$ contradiction.

The only possibility then is that $f(s)=\frac{1}{s}.$ In other words, $$f(x)=\frac{1}{x}$$for all $x\in S.$

We can check that $f(x)=\frac{1}{x}$ works, so $n=1$ and $s=\frac{1}{4},$ so $n\times s=\boxed{\frac{1}{4}}.$