Nice problem. The only functions which work are $f(x)=0$ and $f(x)=x$ for all real $x$. Let $c$ be an arbitrary real and take $P(x)=x-f(c)$. Then $P(f(c))=0$ so $f(P(c))= f(c-f(c))=0$, so $f(x-f(x))=0$ for all real $x$. Now, if $x-f(x)=0$ for all $x$, then $f$ is the identity. Otherwise, suppose $a-f(a)=b$ for some real $a$ and nonzero $b$. Then let $P(x)=rx$ for some arbitrary real $r$; we have that $P(f(a-f(a)))=P(0)=0$ so $f(P(a-f(a)))=f(P(b))=f(rb)=0$. By varying $r$, we conclude that $f$ is everywhere zero. So only the claimed functions work.

Good. $\boxed{f(x)=x}$ is a solution, otherwise assume that $\exists j:f(j)\ne j$. Setting $P(x)=x-f(j)$ and $x=j$, we have $f(j-f(j))=0$. Setting $P(x)=\frac c{j-f(j)}x$ and $x=j-f(j)$, we have $\boxed{f(x)=0}$ for all $x$.

Plugging in $u$ such that $f(u) = 0$ we obtain $f(P(u)) = 0$ for all polynomials $P$ with no constant term. In particular, $f(cu) = 0$ for all $c\in \mathbb{R}$, so if there is some nonzero $u$ with $f(u) = 0$ then $f\equiv 0$. On the other hand, setting $P(y) = y-f(c)$ for some $c$ and setting $x=c$ we obtain $f(c-f(c)) = 0$. Hence if $f(x)\not\equiv x$, then $f(x) = 0$ has a nontrivial solution so the only solutions are $f(x)\equiv x$ and $f(x)\equiv 0$.

This was proposed by me

The only solutions are $\boxed{f(x) \equiv x}$ and $\boxed{f(x) \equiv 0}$. Let $P(x) = a_{n}x^{n} + \cdots + a_{0} = a_{n}(x - r_{1}) \cdots (x-r_{n}) $ where $r_{i}$ is a root of $P$. If $f(x) \equiv 0$ then $P(0) = 0 = a_{n}(x - r_{1}) \cdots (x-r_{n})$ but this implies that there's exists some root $r_{j} = 0$, in other words $P(x) = a_{n}(x - r_{1}) \cdots (x- 0) \cdots (x-r_{n})$ and such $P$ is a polynomial without constant term. To find the other solution $f(x) \equiv x$ you can do a similar proceess as the solutions above and set $P(x) = x - f(k)$ for some $k$, and follow the procedure described above.