Here is an incomplete answer, that hopefully should be useful to conclude in either direction.

For $f:\mathbb{R}\rightarrow\mathbb{R}$ and $P(z):=\sum_{j=0}^n a_j z^j\in\mathbb{R}[z]$ let's denote $f^{(j)}$ the $j$-fold composition and $P[f]:=\sum_{j=0}^n a_j f^{(j)}.$

Here are a few facts about a continuous solution $f:\mathbb{R}\rightarrow\mathbb{R}$ to the functional equation $P[f]=0,$ that should be helpful.

**0.** *If $0$ is not a root of $P$, then $f$ is a homeo, and the inverse of
$f$ satisfies the functional equation $R[f^{-1}]=0$, with the
reciprocal polynomial of $P$, $R(z):=z^nP(1/z)$.*

Indeed, from the equation we have a continuous left-inverse $g:\mathbb{R}\rightarrow\mathbb{R}$ of the form $g:=Q[f]$, so $f$ is injective; but for continuous functions on $\mathbb{R}$, $g\circ f=\mathrm{id_\mathbb{R}}$ also implies that $f$ is unbounded from above and from below, thus surjective.

**1.** *If $1$ is not a root of $P$, then $f$ has no nonzero fixed point. In particular, $f(x)-x$ has constant sign on $\mathbb{R}_+$ and on $\mathbb{R}_-$*

Indeed, $f(x_0)=x_0$ implies again from the equation $P(1)x_0=0$ whence $x_0=0$.

Then, a negative answer to the question would follow from the following statement:

**Lemma (Conjectured)**. Let $(x_m)_{m\ge0}$ be a sequence solution of a linear recurrence $\sum_{j=0}^n a_j x_{m+j}=0,$ with $P(z):=\sum_{j=0}^n a_j
> z^j\in\mathbb{R}[z]$ (and w.l.o.g. $a_na_0\neq0$). Then:

**i)** If $x_m$ has constant sign , then $P$ has a positive real root.

**ii)** If $x_m$ has alternate sign , then $P$ has a negative real root.

If true, a consequence would be:

**2.** *If for some $x_0\neq 0$ the orbit $x_m:=f^{(m)}(x_0)$ has (i) constant sign, then $P$ has a non-negative solution. If (ii) it has alternate sign, then $P$ has a non-positive real root.*

Indeed, again from the equation, the sequence $(x_m)_m$ is a solution of the linear recurrence $\sum_{j=0}^n a_j x_{m+j}=0,$ and both claims follows from the conjectured lemma.

Finally, let's prove how this would imply that $P$ has a real root, assuming that there exists a continuous solution to $P[f]=0$. If $0$ is not a root of $P$, by **(0)** $f$ is either increasing or decreasing. In the first case, all orbits would be monotone, thus with (eventually) constant sign, and $P$ has a positive real root by **(2i)**. In the latter case, $f$ is a decreasing homeomorphism that fixes $0$. But then all orbits have alternate sign, and we conclude by **(2ii)** that $P$ has a negative real root.

*Trying to prove the lemma.* The sequence $(x_m)$ has a representation $x_m:=\sum_{\lambda\in\Lambda}p_\lambda(m)\lambda^m$, where $\Lambda$ is the set of roots of $P$ and $p_\lambda$ are (complex) polynomials with $\mathrm{deg}(P_\lambda)< \mathrm{mult}(\lambda)$. To put in evidence the principal term, this can be written in the form
$x_m:=m^s\rho^m \Big( \sum_{k=1}^r b_k \cos(m\omega_k+\phi_k)+o(1)\Big)$ as $m\to\infty$. If $P$ has no positive solutions, we have $0 < \omega_k<2\pi$, which in several cases implies infinitely many changes of sign for $x_m$ . The claim **(ii)** follows from *(i)*. Indeed, in this case the sequence $y_m:=(-1)^m x_m$ verifies the assumption (i) for the polynomial $P(-z)$, and $P$ has a negative root.