Let $L$ be the splitting field of $Q(x)$ over $\mathbf{Q}(a,b)$. If $\alpha \in L$ is a root of $Q(x)$, then $P(\alpha)$ is also a root of $Q(x)$. Thus the roots of $Q(x)$ are divided up into three pairs:
$$\alpha, P(\alpha), \beta, P(\beta), \gamma, P(\gamma).$$
These are distinct because $Q(x)$ has non-zero discriminant and so is separable. Any automorphism sending $\alpha$ to $\sigma \alpha$ sends $P(\alpha)$ to $P(\sigma \alpha)$, and thus must preserve these three pairs. This puts a restriction on the Galois group $G = \mathrm{Gal}(L/\mathbf{Q}(a,b)) \subset S_6$, namely that it lies inside $\Gamma \subset S_6$, the maximal subgroup of $S_6$ preserving this decomposition.
We shall see that this group is $S_4 \times S_2$.
So far this all generalizes to roots of $P^{(n)}(x) = x$ with $P(x)$ general of degree $d$;
the roots of the corresponding polynomial $Q(x)$ will come in packets of $n$ roots which are iterates under $\alpha \rightarrow P(\alpha)$, and since $\sigma P(\alpha) = P(\sigma \alpha)$, the Galois group will commute with this automorphism, and thus give a restriction on the Galois group as lying inside the centralizer of this permutation.
Here are some more details in the case $(d,n) = (d,2)$. We have
$P^{(2)}(x) = x$ and $Q(x)$ has degree $d^2 - d$, the maximal group $\Gamma$ will preserve the $m:=(d^2-d)/2$ pairs. The largest such group naturally surjects to $S_m$ which describes how it acts on the $m$ pairs. The kernel consists of elements which just permute each pair which is the group $(\mathbf{Z}/2 \mathbf{Z})^m$. This is a group of order $2^m \cdot m!$ and is usually denoted
$$ S_2 \wr S_m =(S_2)^m \ltimes S_m.$$
The symbol on the LHS is the wreath product and this group is called $S_2$ wreath $S_m$; it is defined/identified with the semidirect product preserving $m$ pairs of elements described above. If $\alpha$ is a root of $Q(x)$, the quotient $S_m$ corresponds to considering the Galois group of $\theta = \alpha + P(\alpha)$. You already noticed this when $m = 3$. If you take a random example with $d = 4$, say $P(x)=x^4+x+1$, you will find the Galois group of $Q(x)$ has order $2^6 \cdot 6! = 46080$, whereas if $d = 5$, say $P(x)=x^5+x+1$, it will have order $2^{10} 10! = 3715891200$. In neither case will it be solvable because the groups surject onto $S_6$ and $S_{10}$ respectively.
So far we have seen that $G$ is contained within this group $\Gamma$, but it is also equal to this group, since the group only gets smaller under specialization and one can compute at least one example with $a,b \in \mathbf{Q}$ for which this extension has Galois group $S_4 \times S_2$.
When $d=3$ and so $m=3$ something additional and special happens, namely that there is an isomorphism
$$S_2 \wr S_3 \simeq S_2 \times S_4.$$
This is an exceptional isomorphism which has no analog for higher $d$. You can see that $\Gamma \simeq S_4 \times S_2$ in a number of ways:
Label the pairs of roots which sum to zero as $(1,6)$, $(2,5)$, and $(3,4)$, the condition of being in $\Gamma$ is equivalent to commuting with $\sigma = (1,6)(2,5)(3,4)$. This is because
$$g \sigma g^{-1} = (g(1),g(6))(g(2),g(5))(g(3),g(4)).$$
So it suffices to show that the centralizer of $\sigma$ is $S_4 \times S_2$.
The order of the centralizer is the stabilizer of $G$ acting on $\sigma$ by conjugation. Since $\sigma$ has $15$ conjugates, the order of the stabilizer is $6!/15 = 48$ by the Orbit-Stabilizer theorem. Alternatively the previous description of $\Gamma$ shows that its order is $2^3 \cdot 3! = 48$.
Claim: The centralizer of $\sigma = (16)(25)(34) \in S_6$ is isomorphic to $S_4 \times S_2$.
Proof One: There is an outer automorphism of $S_6$ sending $\sigma$ to a $2$-cycle. The centralizer of $(1,2)$ certainly contains $S_2 \times S_4$ with the first $S_2$ generated by $(1,2)$ and the second $S_4$ acting on $3,4,5,6$. This has order $48$ so we are done.
As a remark, any $S_4 \times S_2$ inside $S_6$ has to normalize an element of order $2$. There are three conjugacy classes of order $2$ elements; $(**)$, $(**)(**)$, and $(**)(**)(**)$. The normalizer of the second conjugacy class has order $6!/45 = 16$, but the first two give the two conjugacy classes of $S_4 \times S_2 \subset S_6$, one that acts transitively and the other that doesn't, but which are interchanged by the exotic automorphism of $S_6$.
Proof Two: Let $C \subset \mathrm{SO}(3)$ denote the group of rotations that preserve the cube (the cube group), with faces labelled in the usual way. It's well known that $C \simeq S_4$ by the action on the $4$ diagonals of the cube. Now $C$ also acts faithfully on the faces, which realizes $S_4 = C$ as a subgroup of $S_6$. Moreover the image commutes with $\sigma$, because $C$ preserves the pairs of opposite faces of the cube. Finally, the reflection $-I \notin C$ both commutes with $C$ and maps to $\sigma$. The group $S_4 \times S_2 \simeq C \times \langle -I \rangle \subset \mathrm{O}(3)$ acts faithfully on the six faces of the cube and also commutes with $\sigma$, and so its image in $S_6$ is isomorphic to $S_4 \times S_2$ and is the centralizer of $\sigma$.
