Let $P(x)$ be a non-zero polynomial and $d = \deg P$. We shall prove that $f(P(x)) = Cd$, for some constant $C$, furthermore $f(0) = 0$. It is easy to check that such a function properly works.
Letting $(P(x), Q(x)) = (P(x), 1)$, $(b, P(x))$ in the second equation as well as $(P(x), Q(x)) = (a, 1)$ in the first equation for some non-zero constants $a, b$ we conclude that $f(a) = 0$ and $f(bP(x)) = f(P(x))$. Letting $(P(x), Q(x)) = (bx, \frac{x}{b}+1)$ to obtain $f(x+b) = f(x+1)$ for all $b \neq 0$. Let $c, d$ be two arbitrary real numbers such that $cd \neq 0$ it follows that
$$f\left(x + \frac{d}{c}\right) = f\left(c\left(x + \frac{d}{c}\right)\right) = f(cx + d) = f(x + 1).$$
Finally, letting $(P(x), Q(x)) = (cx, x - \frac{1}{c})$ for some $c \neq 0, 1$ in the first equation yields
$$f(x) = f(cx) = f\left(cx + 1 - \frac{1}{c}\right) = f(x + 1).$$
Hence, if $P(x)$ is a linear polynomial then $f(P(x)) = f(x+1)$, i.e., a constant function, say $C$. We then finish our proof based on the induction on the degree of polynomial. The base is true for $d=0, 1$. Assume that the statement holds true for all polynomials of degree less than $d$. Choose a non-zero rational number $r$ such that $P(r) \neq 0$. Letting $R(x) = \frac{r}{P(r)}P(x+r)$ it follows that $R(0) = r$. Hence, $R(x) = r+xS(x)$, for some polynomial $S(x)$ of degree $d-1$. Putting $(P(x), Q(x)) = (R(x), x-r)$ in the first equation yielding
$$\begin{aligned} f(R(x) - r) = f(xS(x)) = f(x) + f(S(x)) &= f(R(x-r)) \\ &= f\left(\frac{r}{P(r)}P(x)\right) = f(P(x)). \end{aligned}$$
According to the induction hypothesis $f(S(x)) = C(d-1)$ and $f(x) = C$. Hence,
$$f(P(x)) = Cd.$$