jmc

Setting $x=1$ and $y=-1-f(1),$ we get $$f(f(-1-f(1))+1)=-1-f(1)+f(1)=-1.$$Let $a=f(-1-f(1))+1,$ so $f(a)=-1.$

Setting $y=a,$ we get $$f(0)=ax+f(x).$$Let $b=f(0),$ so $f(x)=-ax+b.$ Substituting into the given functional equation, we get $$-a(x(-ay+b)+x)+b=xy-ax+b.$$This expands as $$a^{2}xy-(ab+a)x+b=xy-ax+b.$$For this to hold for all $x$ and $y,$ we must have $a^2=1,$ and $ab+a=a.$ From $a^2=1,$ $a=1$ or $a=-1.$ For either value, $b=0.$

Hence, the solutions are $f(x)=x$ and $f(x)=-x.$ Therefore, $n=2$ and $s=2+(-2)=0,$ so $n\times s=\boxed{0}.$