jmc

Setting $x=0,$ we get $$f(y+f(0))=f(y)+1$$for all real numbers $y.$

Setting $y=f(0),$ we get $$f(f(x)+f(0))=f(x+f(0))+xf(f(0))-xf(0)-x+1$$for all real numbers $x.$ Since $f(f(x)+f(0))=f(f(x))+1,$ $f(x+f(0))=f(x)+1,$ and $f(f(0))=f(0)+1,$ $$f(f(x))+1=f(x)+1+x(f(0)+1)-xf(0)-x+1.$$This simplifies to $$f(f(x))=f(x)+1.$$Setting $y=0,$ we get $$f(f(x))=f(x)+xf(0)-x+1.$$But $f(f(x))=f(x)+1,$ so $xf(0)-x=0$ for all $x.$ This means $f(0)=1.$ Hence, $$f(x+1)=f(x)+1$$for all $x.$

Replacing $x$ with $x+1,$ we get $$f(f(x+1)+y)=f(x+y+1)+(x+1)f(y)-(x+1)y-x+1.$$Since $f(f(x+1)+y)=f(f(x)+y+1)=f(f(x)+y)+1$ and $f(x+y+1)=f(x+y),$ we can write this as $$f(f(x)+y)+1=f(x+y)+1+(x+1)f(y)-(x+1)y-x+1.$$Subtracting $f(f(x)+y)=f(x+y)+xf(y)-xy-x+1,$ we get $$1=f(y)-y,$$so $f(x)=x+1$ for all $x.$ We can check that this function works.

Therefore, $n=1$ and $s=2,$ so $n\times s=\boxed{2}.$