$f(x) = x + c$ for some constant $c$.

We first check that all such functions are solutions. Indeed, if $f(x) = x + c$, then
$$f(x + f(y)) = f(x + y + c) = x + y + 2c = f(f(x)) + y$$
for any $x, y \in \mathbb{R}$.

Next, we prove that these are the only solutions. Replacing $x$ by $x - f(y)$, the given condition means
$$f(x) = x \quad \text{or} \quad f(x) = f(f(x - f(y))) + y$$
for any $x, y \in \mathbb{R}$. We are done if $f(x) = x$ for all $x \in \mathbb{R}$. Assume $f(a) \neq a$ for some $a \in \mathbb{R}$. Then we have
$$f(a) = f(f(a - f(y))) + y \quad (1)$$
for any $y \in \mathbb{R}$. Note that $f$ is injective, since if $f(\alpha) = f(\beta)$, then
$$\alpha = f(\alpha) - f(f(\alpha - f(\alpha))) = f(\alpha) - f(f(\alpha - f(\beta))) = \beta$$
by (1). Setting $y = 0$ in (1), we have $f(a) = f(f(a - f(0)))$, and hence
$$a = f(a - f(0)) \quad (2)$$
as $f$ is injective. In particular, this implies $f(0) \neq 0$ (as $f(a) \neq a$).

* If $f(x) \neq x$ for all $x \in \mathbb{R}$, then (2) holds for all $a \in \mathbb{R}$. Replacing $a$ by $x + f(0)$, this gives $f(x) = x + c$ for all $x \in \mathbb{R}$, where $c = f(0)$ is a constant as desired.

* Suppose $f(b) = b$ for some $b$. Note that $f(f(a)) \neq f(a)$ since $f(a) \neq a$ and $f$ is injective. It follows that (1) holds if we replace $a$ by $f(a)$. Thus, we obtain
$$f(f(a)) = f(f(f(a) - f(y))) + y.$$
Putting $y = a$, this yields
$$f(f(a)) = f(f(0)) + a. \quad (3)$$
Next, putting $y = b$ in (1), we have $f(a) = f(f(a - b)) + b$. Recall that $f(a) \neq a$, so we cannot have $f(a - b) = a - b$. Therefore, using (3) with $a$ replaced by $a - b$, we deduce
$$f(a) = f(f(a - b)) + b = f(f(0)) + (a - b) + b = a + f(f(0)). \quad (4)$$
Recall that $f(0) \neq 0$. Thus, we may take $a = 0$ in (4). This implies $f(0) = f(f(0))$. But then this yields $0 = f(0)$ by the injectivity, which is a contradiction.