There are many many solutions. Let $S\subseteq\mathbb R$ be any nonempty subset satisfying $2S\subseteq S$. Pick any function $g:\mathbb R\setminus S\to S$. Let $$ f(x)=\begin{cases} x&\text{if }x\in S\\ g(x)-x&\text{otherwise.} \end{cases} $$ For each $x\in\mathbb R$ we have $f(x)+x\in S$, so $f(f(x)+x)=f(x)+x$.

There are many many solutions. Let $S\subseteq\mathbb R$ be any nonempty subset satisfying $2S\subseteq S$. Pick any function $g:\mathbb R\setminus S\to S$. Let $$ f(x)=\begin{cases} x&\text{if }x\in S\\ g(x)-x&\text{otherwise.} \end{cases} $$ For each $x\in\mathbb R$ we have $f(x)+x\in S$, so $f(f(x)+x)=f(x)+x$.

Even if we require $f$ to be continuous, there are as many solutions as continuous functions on $\mathbb R$. Indeed set $S=\{x\in\mathbb R\mid x\geq0\}$ and pick any continuous function $g:\mathbb R\setminus S\to S$ satisfying $\lim_{x\to0^-}g(x)=0$; then construct $f$ as above.