jmc

Setting $x=-f(y),$ we get $$f(0)=-f(y)+y,$$so $f(y)=y-f(0)$ for all real numbers $x.$ Then the given functional equation becomes $$f(x+y-f(0))=x+y,$$or $x+y-f(0)-f(0)=x+y.$ Then $f(0)=0,$ so $f(x)=x$ for all real numbers $x.$ This function does satisfy the given functional equation, giving us $\boxed{1}$ solution.