The elements of $G=\text{Gal}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q})$ can be identified with permutations of the three objects $a=\sqrt[3]{2}$, $b=\zeta_3\sqrt[3]{2}$, and $c=\zeta_3^2\sqrt[3]{2}$ as follows: An element $g\in G$ permutes these 3 objects by $a\mapsto g(a)$ $b\mapsto g(b)$ $c\mapsto g(c)$ and furthermore $g$ is uniquely determined by how it permutes these three objects.
Thus, each $g\in G$ can be associated with a permutation $\sigma_g\in S_3$ in such a way that the map $f:G\rightarrow S_3$ defined by $f(g)=\sigma_g$ is an injective homomorphism. We then identify $G$ with its image in $S_3$, i.e. we forget about the distinction between them - we can do this because $f$ is injective, so there is no "loss of information". This is the sense in which $G$ "is a subgroup" of $S_3$.