Let $G_n$ denote the subgroup of the orthogonal group $O_n$ of elements that send the hypercube to itself, the group of symmetries $C_n$, including the orientation-reversing symmetries.
It would like to show that $G_2$ is isomorphic to the dihedral group $D_4$ of order 8
I determined a formula for the order of $G_n$ which is $n!2^n$. Therefore, we already notice that both sets have the same order.
OK, we have established that your definition of $D_4$ is $$ D_4 = \langle a, b \mid a^4 = b^2 = 1, b^{-1} a b = a^{-1} \rangle. $$
Now, in the group $G_2$ choose $A$ to be a rotation by 90 degrees and $B$ to be a mirror reflection. Clearly, there are such $A$ and $B$ in $G_2$. It is also clear that $G_2$ is generated by $A$ and $B$.
To prove that $G_2$ is isomorphic to $D_4$, we will build homomorphisms $\varphi: D_4 \to G_2$ and $\psi: G_2 \to D_4$ that are inverses of each other.
To build $\varphi: D_4 \to G_2$, let us note that in group $G_2$ elements $A$ and $B$ satisfy these relations: $A^4 = 1$, $B^2 = 1$, $B^{-1} A B = A^{-1}$ (check this by hand). It follows that there exists a unique homomorphism $\varphi: D_4 \to G_2$ that sends $a$ to $A$ and $b$ to $B$.
Now let us build $\psi: G_2 \to D_4$. Note that every element $C \in G_2$ can be represented as $C = A^m B^n$. Define $\psi(C) = a^m b^n$.
First of all, we need to check that this definition is correct. If we represent $C$ in two different ways, $C = A^m B^n = A^{m'} B^{n'}$, we need to check that the same value is assigned to $\psi(C)$, i.e. $a^m b^n = a^{m'} b^{n'}$. Indeed, if $A^m B^n = A^{m'} B^{n'}$, then $A^{m-m'} = B^{n'-n}$. $A$ preserves orientation and $B$ doesn't. It follows that $n-n'$ is even, so $A^{m-m'}=1$, so $m-m'$ is a multiple of $4$. But then $$a^{m'}b^{n'} = a^m a^{m'-m} b^n b^{n'-n} = a^m b^n,$$ because $a^4 = b^2 = 1$ in $D_4$. So map $\psi$ is defined correctly.
We also need to check that $\psi$ is a homomorphism. This is quite mundane and involves a little case analysis, so I will omit the check.
Now that we have built $\psi$ and $\varphi$, we see that $\psi\varphi(a) = a$ and $\psi\varphi(b) = b$, so $\psi\varphi$ is the identity map of $D_4$. In a similar way $\varphi\psi$ is the identity map of $G_2$. It follows that $\psi$ and $\varphi$ are inverses of each other, so they are isomorphisms, qed.