This is a summary of my calculations based on the comments by David E. Speyer. I used a different description of the Galois group, but that doesn't matter.
The basic tool in use (see the earlier comments) is Dedekind's theorem relating factorizations of
$$
f(x)= x^6 + 2 x^5 + 3 x^4 + 4 x^3 + 5 x^2 + 6 x + 7
$$
modulo various (unramified) primes to the cycle structure of elements of the Galois group $G$ as permutations of the roots (here six, so $G\le S_6$). As explained by the OP
- Modulo $p=3$ $f(x)$ is irreducible, so there is a 6-cycle in $G$. In particular $G$ is transitive and $f(x)$ is also irreducible over $\Bbb{Q}$.
- Modulo $p=11$ $f(x)$ splits as a product of a linear and a quintic, implying that $G$ contains a 5-cycle. Therefore the point stabilizer of $G$ among the set of roots acts transitively among the remaining roots, and $G$ is doubly transitive.
- The testing I did revealed that modulo $p=19$ $f(x)$ splits as a product of two linear factors and an irreducible quartic. Therefore $G$ contains a 4-cycle, and we can also conclude that $G$ is a triply transitive subgroup of $S_6$.
- More testing only gave cycle structures that are powers of the already listed, so this suggested that $G$ might be a transitive copy of $S_5$ inside $S_6$. It turns out that this is true. Let's record the fact that we already know $G$ to have order at least $120$.
I constructed that copy of $S_5$ from its conjugation action on its six Sylow $5$-groups:
$P_1=\langle(12345)\rangle$, $P_2=\langle(12354)\rangle$, $P_3=\langle(12435)\rangle$,
$P_4=\langle(12453)\rangle$, $P_5=\langle(12534)\rangle$ and $P_6=\langle(12543)\rangle$.
The group $S_5$ is generated by $\alpha=(12345)\in P_1$ and $\beta=(45)$. Conjugation action by $\alpha$ permutes the subscripts of the $P_i$s according to $a=(24653)$ whereas
conjugation by $\beta$ permutes the Sylow-$5$s according to $b=(12)(35)(46)$. It follows that a desired copy of $S_5$ is the group
$$
\tilde{G}=\langle(24653),(12)(35)(46)\rangle.
$$
Using a suitable CAS (I used Mathematica) it is then straight forward to generate a list of elements of $\tilde{G}$.
David E. Speyer's idea is that the polynomial
$$
P(x_1,x_2,\ldots,x_6)=\sum_{g\in \tilde{G}}x_{g(1)}^2x_{g(2)}^2x_{g(3)}x_{g(4)}
$$
is invariant under $\tilde{G}$ but isn't invariant under $S_6$. Because $\tilde{G}$ is a maximal subgroup of $S_6$ we can use $P$ to distinguish the Galois groups $\tilde{G}$ and
$S_6$. Namely, if $P$ evaluated at a carefully chosen permutation of the roots of $f(x)$
produces an integer, then that equation must be respected by the Galois group $G$, implying $G=\tilde{G}$.
Mathematica kindly gave me approximate zeros, and those are (rounded to only 4 decimals to save space)
$$
\begin{aligned}
z_1\approx-1.3079-0.5933i,&&z_2=\overline{z_1},\\
z_3\approx-0.4025-1.3417i,&&z_4=\overline{z_3},\\
z_5\approx\hphantom{-}0.7104-1.1068i,&&z_6=\overline{z_5}.
\end{aligned}
$$
Which permutation of these roots should we use? Because $\tilde{G}$ is triply transitive,
there is no need to try anything other than $x_1=z_1,x_2=z_2,x_3=z_3$. Simply try out the different orderings of $z_4,z_5,z_6$. It turns out that
$$
P(z_1,z_2,z_3,z_6,z_5,z_4)=264
$$
an integer to the precision I had available.
At this point I also tested that "complex conjugation" (based on the positions of the conjugate pairs in the list of variables), i.e. the permutation
$(12)(36)(45)$ is, indeed, an element of $\tilde{G}$. This added to my confidence :-)
This could still be a false alarm in the sense that the actual value might just happen to be extremely close to $264$. David's suggestion was to look at the polynomial
$$
H(T)=\prod_{\sigma\in Sym\{4,5,6\}}(T-P(z_1,z_2,z_3,z_{\sigma(4)},z_{\sigma(5)},z_{\sigma(6)})).
$$
This is known to be invariant under $S_6$, and hence absolutely guaranteed to have integer coefficients. The idea is to verify that $H(264)=0$, which we can do with exact integer arithmetic alone. An expansion (still using approximate zeros, but now "legally" allowed to round the coefficients of $H(T)$ to the obvious nearest integer) gives
$$
H(T)=T^6-240 T^5-101440 T^4+24410112 T^3+2093608960 T^2-447570968576 T-1492648329216,
$$
and we can readily check that $H(264)=0$.
