Let the function $f$ is such that for every $x, y \in \mathbb{R}$ it is fulfilled the equation (1).
If in (1) we put $y = x$, we obtain
$$f(x + f(x)) = 2x, \text{ for every } x \in \mathbb{R}. \quad (2)$$
Furthermore, for $x = 0$ from (2) follows $f(f(0)) = 0$, and for $x = y = \frac{f(0)}{2}$ from (1) we obtain $f(f(0)) = f(0)$. Now, from the last two equations it follows that $f(0) = 0$.
Now, if in (1) we put $y = 0$ and if we use $f(0) = 0$, we obtain
$$f(\max\{x, 0\} + \min\{f(x), 0\}) = x, \text{ for every } x \in \mathbb{R}. \quad (3)$$
We will consider two cases:
Case 1. $x > 0$. Then from (3) it follows
$$f(x + \min\{f(x), 0\}) = x.$$
From the last equality follows that or $f(x + f(x)) = x$ or $f(x) = x$. If $f(x + f(x)) = x$, then for (2) it follows $2x = x$, which is not possible when $x > 0$. Hence, $f(x) = x$.
**Case $2^\circ$** $x < 0$. Then from (3) it follows
$$f(\min\{f(x), 0\}) = x.$$
Now, from the last equality we obtain or $f(0) = x$ or $f(f(x)) = x$. If $f(0) = x$, then $0 > x = f(0) = 0$, which is a contradiction. Hence, $f(f(x)) = x$. Now, if in (2) instead of $x$ we put $f(x)$ and if first we use that $f(f(x)) = x$, and after that we use the equality (2), we obtain
$$2f(x) = f(f(x) + f(f(x))) = f(f(x) + x) = 2x,$$
from where we obtain $f(x) = x$, for $x < 0$.
Finally, $1^\circ$ and $2^\circ$ implies that $f(x) = x$ for every $x \in \mathbb{R}$. It is not difficult to check that this function is a solution of the problem.