Your description of $G$ is perfectly fine as it is.
But maybe a representation of $G$ in $GL_2(\mathbb{Z}/8\mathbb{Z})$, would be more to your taste :
if $\sigma \in G$ satisfies $\sigma(\zeta_8) = \zeta_8^a$ and $\sigma(\sqrt[8]{3}) = \zeta_8^b \sqrt[8]{3}$, it is represented by the matrix
$ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}$
You can observe that the automorphisms fixing $\sqrt[8]{3}$ are those with $b=0$, and they form a (non normal) subgroup $H$ isomorphic to $(\mathbb{Z}/8\mathbb{Z})^*$.
The automorphisms fixing $\zeta_8$ are those with $a=1$, and they form a normal subgroup $N$ isomorphic to $\mathbb{Z}/8\mathbb{Z}$, and $G$ can also be seen as a semi-direct product of those two groups :
In the exact sequence
$0 \rightarrow N = Gal_{\mathbb{Q}(\zeta_8)}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) \rightarrow G = Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) \rightarrow G/N = Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8)) \rightarrow 0$
there is a section $s : Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8)) \rightarrow H \subset Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) $, defined simply by extending an automorphism $\sigma$ with $s(\sigma)(\sqrt[8]{3}) = \sqrt[8]{3}$.
$Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8))$ is isomorphic to $(\mathbb{Z}/8\mathbb{Z})^*$, so that the canonical surjection $f$ is simply picking $f(\sigma) = a$, and the section $s : (\mathbb{Z}/8\mathbb{Z})^* \rightarrow H$,
is $s(a) = \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$
This gives an isomorphism between $G/N$ and $H$ and shows that $G$ is the semidirect product of $H$ acting on $N$
Not every permutation of the roots is an automorphism of $E/\mathbb Q$. For example, any permutation $\sigma$ which sends $\sqrt2$ to $\sqrt3$ is not an automorphism, since $\sigma(\sqrt2^2)=\sigma(2)=2$ but $\sigma(\sqrt2)^2=3$.
More generally any automorphism of $E/\mathbb Q$ must preserve the roots of polynomials with rational coefficients, in particular the roots of $x^2-2$ and $x^2-3$. So the only possible automorphisms are given by
$$\sqrt2\mapsto \sqrt2,\ \sqrt3\mapsto \sqrt3$$
$$\sqrt2\mapsto -\sqrt2,\ \sqrt3\mapsto \sqrt3$$
$$\sqrt2\mapsto \sqrt2,\ \sqrt3\mapsto -\sqrt3$$
$$\sqrt2\mapsto -\sqrt2,\ \sqrt3\mapsto -\sqrt3$$
and since there are exactly $4$ automorphisms, you know that all of these are automorphisms.
Best Answer
You are correct in saying that the splitting field can be written $\Bbb{Q}(a,i)$.
However there are three subgroups of order four.
1: The cyclic subgroup
The automorphism group contains a cycle of the roots, namely the map that takes $a$ to $\frac{i}{a}$ and $i$ to $-i$, which corresponds to the cycle $$\pmatrix{a & \frac{i}{a} & -a & \frac{-i}{a}}.$$
Now the subfield fixed by the group generated by this automorphism has degree 2 over $\Bbb{Q}$ by the main theorem of Galois theory. Note in addition that $ia^2+\frac{i}{a^2}$ is fixed by this map since $ia^2\mapsto -i(\frac{i^2}{a^2})=\frac{i}{a^2}$, and $\frac{i}{a^2}\mapsto -i(\frac{a^2}{i^2})=ia^2$. However $$ia^2+\frac{i}{a^2}=i(1+\sqrt{2})+\frac{i}{1+\sqrt{2}}=i\left(1+\sqrt{2}+\frac{1-\sqrt{2}}{1-2}\right)=2i\sqrt{2},$$ which has degree two over $\Bbb{Q}$. Thus the fixed field corresponding to this subgroup is $$\Bbb{Q}\left(ia^2+\frac{i}{a^2}\right)=\Bbb{Q}[2i\sqrt{2}]=\Bbb{Q}[\sqrt{-2}].$$
2: A Klein 4 subgroup
Another subgroup of order four is generated by the automorphism that sends $a$ to $-a$ and $i$ to $i$ and the automorphism that sends $a$ to $a$ and $i$ to $-i$. Note that both of these fix $a^2$, which has order 2 over $\Bbb{Q}$, so the fixed field corresponding to this subgroup is $$\Bbb{Q}[a^2]=\Bbb{Q}[1+\sqrt{2}]=\Bbb{Q}[\sqrt{2}].$$
3: Another Klein 4 subgroup
Finally, there must be a third subgroup of order four, which is generated by the automorphism that sends $a$ to $\frac{i}{a}$ and $i$ to $i$ and the automorphism that sends $a$ to $-a$ and $i$ to $i$. Note that these both fix $i$, so the fixed field is $\Bbb{Q}[i]$.
Note that this makes a lot of sense, subgroups of order 4 correspond to fixed fields of degree 2 over the base field, or quadratic extensions. Therefore, we expect $\Bbb{Q}[i]$ and $\Bbb{Q}[\sqrt{2}]$, since both of these are clearly contained in the splitting field. Then we notice that since $\Bbb{Q}[\sqrt{2}]$ is a real field, and since $\Bbb{Q}[i]$ contains no real irrational elements, $\Bbb{Q}[i\sqrt{2}]=\Bbb{Q}[\sqrt{-2}]$ must be the last field (if you count subgroups of $D_4$, you can check there are 3 of order 4). Then we just have to match these fields up to groups.