jmc

Setting $x=y=1,$ we get $$f(1{)}^2-f(1)=2,$$so $f(1{)}^2-f(1)-2=0.$ This factors as $(f(1)+1)(f(1)-2)=0,$ so $f(1)=-1$ or $f(1)=2.$

Setting $y=1,$ we get $$f(x)f(1)-f(x)=x+1$$for all $x.$ Then $f(x)(f(1)-1)=x+1.$ Since $f(1)\ne 1,$ we can write $$f(x)=\frac{x+1}{f(1)-1}.$$If $f(1)=-1,$ then $$f(x)=\frac{x+1}{-2},$$and we can check that this function does not work.

If $f(1)=2,$ then $$f(x)=x+1$$and we can check that this function works.

Therefore, $n=1$ and $s=3,$ so $n\times s=\boxed{3}.$