I'm not sure if we can derive all analytic solutions to this functional equation outside of polynomials, but we can prove that in polynomial space the only possibilities are $f(x) = x$ and $f(x) = -x$
First, let's assume that $f(x) = ax + b$. Then, we have:
$f(x + f(x)) = a(x + f(x)) + b = ax + b + af(x) = f(x) + af(x) = (a+1)f(x)$
Therefore, $f(x + f(x)) = x + f(x) = (a + 1)f(x)$, which means that $x = a \cdot f(x)$
Then, $x = a(ax + b)$, and thus $(a^2-1)x + ab = 0$
This gives us a system of equations: $a^2 - 1 =0$ and $ab=0$
From the first one, $a = \pm 1$, which means $a \neq 0$ and this forces $b = 0$; therefore the only linear solutions to the equation are $f(x) = x$ and $f(x) = -x$.
Now, suppose that there are polynomial solutions of degree higher than $1$. This means we can write $f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + ... a_nx^n$.
Then, $f(x + f(x))$ can be written as $a_0 + a_1(x + f(x)) + a_2(x + f(x))^2 + a_3(x + f(x))^3 + ... a_n(x + f(x))^n$
Now, focus on that last term: $a_n(x + f(x))^n$. The largest term in the polynomial expansion of this term has a degree of $n^2$. However, the degree of $x + f(x)$ is just the degree of $f(x)$, which is $n$; therefore, the only possible solutions happen when these degrees are equal, or simply: $n^2 = n$, which gives the solutions $n = 0$ and $n = 1$, covered by the prior part of this solution. Any degree higher than 1 and we don't have any solutions.
We cannot, unfortunately, expand this idea to cover infinite power series, so I can't provide any other obvious solutions to the equation.