Clearly, each of the functions $x \mapsto x$ and $x \mapsto 2-x$ satisfies (1). It suffices now to show that they are the only solutions to the problem. Suppose that $f$ is any function satisfying (1). Then setting $y=1$ in (1), we obtain
$$f(x+f(x+1))=x+f(x+1)$$
in other words, $x+f(x+1)$ is a fixed point of $f$ for every $x \in \mathbb{R}$. We distinguish two cases regarding the value of $f(0)$.

Case 1. $\quad f(0) \neq 0$. By letting $x=0$ in (1), we have
$$f(f(y))+f(0)=f(y)+y f(0)$$
So, if $y_{0}$ is a fixed point of $f$, then substituting $y=y_{0}$ in the above equation we get $y_{0}=1$. Thus, it follows from (2) that $x+f(x+1)=1$ for all $x \in \mathbb{R}$. That is, $f(x)=2-x$ for all $x \in \mathbb{R}$.

Case 2. $\quad f(0)=0$. By letting $y=0$ and replacing $x$ by $x+1$ in (1), we obtain
$$f(x+f(x+1)+1)=x+f(x+1)+1$$
From (1), the substitution $x=1$ yields
$$f(1+f(y+1))+f(y)=1+f(y+1)+y f(1)$$
By plugging $x=-1$ into (2), we see that $f(-1)=-1$. We then plug $y=-1$ into (4) and deduce that $f(1)=1$. Hence, (4) reduces to
$$f(1+f(y+1))+f(y)=1+f(y+1)+y$$
Accordingly, if both $y_{0}$ and $y_{0}+1$ are fixed points of $f$, then so is $y_{0}+2$. Thus, it follows from (2) and (3) that $x+f(x+1)+2$ is a fixed point of $f$ for every $x \in \mathbb{R}$; i.e.,
$$f(x+f(x+1)+2)=x+f(x+1)+2$$
Replacing $x$ by $x-2$ simplifies the above equation to
$$f(x+f(x-1))=x+f(x-1)$$
On the other hand, we set $y=-1$ in (1) and get
$$f(x+f(x-1))=x+f(x-1)-f(x)-f(-x)$$
Therefore, $f(-x)=-f(x)$ for all $x \in \mathbb{R}$. Finally, we substitute $(x, y)$ by $(-1,-y)$ in (1) and use the fact that $f(-1)=-1$ to get
$$f(-1+f(-y-1))+f(y)=-1+f(-y-1)+y$$
Since $f$ is an odd function, the above equation becomes
$$-f(1+f(y+1))+f(y)=-1-f(y+1)+y$$
By adding this equation to (5), we conclude that $f(y)=y$ for all $y \in \mathbb{R}$.