Very nice. I'll prove either $f(x)=x$ for all $x$ or $f(x)=-2x$ for all $x$. Denote by $P(x,y)$ the given assertion. First, $P(-f(y),y)$ gives $f(y) = f(-f(y)) + 2y$. In particular, $f$ is injective. Next, $P(x,0)$ gives $f(x+f(0)) = f(x)$, which, coupled with injectivity, gives $f(0)=0$. Moreover, $P(x,-x)$ gives $f(f(-x)) = f(x)-2x$. As $f(y) -2y= f(-f(y))$ and $f$ is injective, we get that $f$ is odd: $f(-x)=-f(x)$. Also, $P(0,y)$ gives $f(y+f(y)) = 2y$ and inserting $y+f(y)$ in place of $y$ in $f(y) = 2y-f(f(y))$, we find $2y = 2(y+f(y)) -f(2y)$. So, $f(2y)=2f(y)$. In particular if $f(1)=a$ then $f(2)=2a$. Lastly, $P(x,f(y))$ gives $f(x+2y) = f(x)+f(2y)$. From here, we iterate to find $f(x) = xa$ for any $x$. Solving, we get $a\in\{1,-2\}$, which yields the solutions claimed above.