jmc

Setting $y=0,$ we get $$f(f(x))=xf(0)+f(x)$$for all $x.$ In particular, $f(f(0))=f(0).$

Setting $x=f(0)$ and $y=0,$ we get $$f(f(f(0)))=f(0{)}^2+f(f(0)).$$Note that $f(f(f(0)))=f(f(0))=f(0)$ and $f(f(0))=f(0),$ so $f(0)=f(0{)}^2+f(0).$ Then $f(0{)}^2=0,$ so $f(0)=0.$ It follows that $$f(f(x))=f(x)$$for all $x.$

Setting $x=1$ in the given functional equation, we get $$f(y+1)=f(y)+1$$for all $y.$ Replacing $y$ with $f(x),$ we get $$f(f(x)+1)=f(f(x))+1=f(x)+1.$$For nonzero $x,$ set $y=\frac{1}{x}$ in the given functional equation. Then $$f(1+f(x))=xf\left(\frac{1}{x}\right)+f(x).$$Then $xf\left(\frac{1}{x}\right)+f(x)=f(x)+1,$ so $xf\left(\frac{1}{x}\right)=1,$ which means $$f\left(\frac{1}{x}\right)=\frac{1}{x}$$for all $x\ne 0.$

We conclude that $f(x)=x$ for all $x.$ Therefore, $n=1$ and $s=\frac{1}{2},$ so $n\times s=\boxed{\frac{1}{2}}.$