Let $z_1, z_2$ be real numbers for which $f(z_1) = f(z_2)$. Substituting $y = z_1$ and $y = z_2$ into the given equation we get $f(f(x) + f(z_1)) = f(x) + z_1$, and $f(f(x) + f(z_2)) = f(x) + z_2$. Since the left hand sides are equal, we have $f(x) + z_1 = f(x) + z_2$, whence $z_1 = z_2$. Hence $f$ is one-to-one. Substituting $y = 0$ into the given equation we get $f(f(x) + f(0)) = f(x)$ for any real $x$. Since $f$ is one-to-one, we have $f(x) + f(0) = x$. If $x = 0$, then the last equation gives $2f(0) = 0$, or $f(0) = 0$. So this equation simplifies to $f(x) = x$. The function $f(x) = x$ satisfies the original equation.

Interchanging $x$ and $y$ in the given equation we get $f(f(y) + f(x)) = f(y) + x$. Since the left hand side is the same as in the original equation, we have $f(x) + y = f(y) + x$. Substituting $y = 0$ into this we get $f(x) = x + f(0)$. Substituting into the original equation all applications of $f$ according to the last equality, we get $x + f(0) + y + f(0) + f(0) = x + f(0) + y$. This gives $f(0) = 0$ and from $f(x) = x + f(0)$ we get $f(x) = x$.