jmc

Setting $x=y=0,$ we get $$f(f(0))=2f(0).$$Let $c=f(0),$ so $f(c)=2c.$

Setting $x=0$ and $y=c,$ we get $$f(0)=f(0)+f(f(c)-c).$$Then $f(c)=0,$ so $c=0.$

Setting $x=0,$ we get $$f(-y)=f(f(y))$$for all $y.$

Setting $y=f(x),$ we get $$0=f(x)+f(f(f(x))-f(-x))+x.$$Since $f(f(x))=f(-x),$ this becomes $f(x)=-x$ for all $x.$ We can check that this function works.

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