The answer is yes, $P$ must be constant. By $S(n)$ we mean the sum of the decimal digits of $|n|$.
We need two claims.

Claim — If $P(x) \in \mathbb{Z}[x]$ is nonconstant with positive leading coefficient, then there exists an integer polynomial $F(x)$ such that all coefficients of $P \circ F$ are positive except for the second one, which is negative.
Proof. We will actually construct a cubic $F$. We call a polynomial good if it has the property.
First, consider $T_0(x) = x^3 + x + 1$. Observe that in $T_0^{\text{deg } P}$, every coefficient is strictly positive, except for the second one, which is zero.
Then, let $T_1(x) = x^3 - \frac{1}{D}x^2 + x + 1$. Using continuity as $D \to \infty$, it follows that if $D$ is large enough (in terms of $\text{deg } P$), then $T_1^{\text{deg } P}$ is good, with $-\frac{3}{D}x^{3\text{deg } f-1}$ being the only negative coefficient.
Finally, we can let $F(x) = CT_1(x)$ where $C$ is a sufficiently large multiple of $D$ (in terms of the coefficients of $P$); thus the coefficients of $(CT_1(x))^{\text{deg } P}$ dominate (and are integers), as needed. $\square$

Claim — There are infinitely many Fibonacci numbers in each residue class modulo 9.
Proof. Easy. First note the Fibonacci sequence is periodic modulo 9 (indeed it is periodic modulo any integer). Moreover (allowing negative indices),
$$\begin{align*} F_0 &= 0 \equiv 0 \pmod{9} \\ F_1 &= 1 \equiv 1 \pmod{9} \\ F_3 &= 2 \equiv 2 \pmod{9} \\ F_4 &= 3 \equiv 3 \pmod{9} \\ F_7 &= 13 \equiv 4 \pmod{9} \\ F_5 &= 5 \equiv 5 \pmod{9} \\ F_{-4} &= -3 \equiv 6 \pmod{9} \\ F_9 &= 34 \equiv 7 \pmod{9} \\ F_6 &= 8 \equiv 8 \pmod{9}. \end{align*} \square$$

We now show how to solve the problem with the two claims. WLOG $P$ satisfies the conditions of the first claim, and choose $F$ as above. Let
$$P(F(x)) = c_N x^N - c_{N-1} x^{N-1} + c_{N-2} x^{N-2} + \dots + c_0$$
where $c_i > 0$ (and $N = 3 \text{deg } P$). Then if we select $x = 10^e$ for $e$ large enough (say $x > 10 \max_i c_i$), the decimal representation $P(F(10^e))$ consists of the concatenation of
* the decimal representation of $c_N - 1$,
* the decimal representation of $10^e - c_{N-1}$
* the decimal representation of $c_{N-2}$, with several leading zeros,
* the decimal representation of $c_{N-3}$, with several leading zeros,
* ...
* the decimal representation of $c_0$, with several leading zeros.
(For example, if $P(F(x)) = 15x^3 - 7x^2 + 4x + 19$, then $P(F(1000)) = 14,993,004,019$.)
Thus, the sum of the digits of this expression is equal to
$$S(P(F(10^e))) = 9e + k$$
for some constant $k$ depending only on $P$ and $F$, independent of $e$. But this will eventually
hit a Fibonacci number by the second claim, contradiction.