The answer by Lukas Heger of course settles all finite groups at once.
Still, for $S_3$ it is very possible to give an explicit hands-on description of this map.
Such an isomorphism is essentially nothing more than writing out the decomposition of $\mathbb{C}[S_3]$ into primary components explicitly. This is not too difficult to do: For notation, let $\tau$ be any nontrivial 3-cycle (so that $A_3=\{1,\tau, \tau^2\}$) and $\sigma$ be any transposition.
The subspace of fixed points of $\mathbb{C}[S_3]$ is just $T$, the line of all vectors with constant coefficients, i.e. all multiples of the vector $\sum_{g\in S_3}g$.
Similarly, the subspace $S$ corresponding to the sign representation can be easily determined to be all multiples of the vector
$$1+\tau+\tau^2-\sigma(1+\tau+\tau^2),$$
i.e. all those vectors that have equal coefficients on any coset of $A_3$ but overall coordinate sum of $0$. (This can for instance be determined from the observation that both the sign and the trivial representation are fixed by $A_3$ - hence both components must lie in the subspace of $A_3$-fixed points, which are just the multiples of $1+\tau+\tau^2$.)
Thus, the last remaining component is the four dimensional invariant complement of $T\oplus S$, which is just
$$V=\left \{ x\in \mathbb{C}[S_3]:\sum_{g\in A_3}x_g g=\sum_{g\in \sigma A_3}x_g g=0 \right \}. $$
These three subspaces $T,S$ and $V$ are now exactly your desired $\mathbb{C}, \mathbb{C}$ and $M_2(\mathbb{C})$. The isomorphism from $\mathbb{C}[S_3]$ can then be read off from the projection onto these subspaces (i.e. primary idempotents). These can be cleanly written in terms of $u=\frac{1}{3}(1+\tau+\tau^2)$: The idempotent in $T$ is $e_T=\frac{u+\sigma u}{2}$, in $S$ we have $e_S=\frac{u-\sigma u}{2}$ and for $V$ we have the complementary
$$e_V=1-e_T-e_S=1-u.$$
Altogether this explicitly determines the algebra isomorphism
$$\mathbb{C}[S_3]\to T\times S\times V, x\mapsto (e_Tx,e_Sx,e_Vx).$$
The further identifications $T\simeq S\simeq \mathbb{C}$ are of course canonical, while $V\simeq M_2(\mathbb{C})$ is determined only up to isomorphism, which is essentially a choice on how to split $V$ further into two complementary $S_3$-invariant subspaces. I will leave it to you to figure out a specific such isomorphism.
Best Answer
If $\mathbb F$ is any field of characteristic zero, then $\mathbb F[D_8]$ is a direct sum of matrix algebras over division rings since by Maschke's theorem it is semisimple. Now as the OP points out $$ D_8/[D_8,D_8] = D_8/Z(D_8)\cong \mathbb Z_2\times \mathbb Z_2=V_4 $$ where $V_4$ is called the "Klein 4-group", and it is easy to see that $\mathbb Q[V_4] \cong \mathbb Q^4$ (the isomorphism is just writing down all the 1-dimensional representations).
Now $Z(\mathbb Q[D_8])$ is $5$-dimensional, since $D_8$ has 5 conjugacy classes, and five irreducible representations, hence either $\mathbb Q[D_8]$ is either $\mathbb Q^4 \oplus \mathrm{Mat}_2(\mathbb Q)$ or $\mathbb Q[D_8] \cong \mathbb Q^4 \oplus D$ where $D$ is a $4$-dimensional noncommutative division algebra, and to decide which, you can just write down the two-dimensional representation $V$ and see what the image of $D_8$ spans inside $\text{End}(V)$: Indeed $$ D_8 = \langle s,r\mid s^4=r^2=e, srs^{-1}=r^{-1}\rangle $$ and the two-dimensional representation is given by $$ r \mapsto \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right); \quad s \mapsto \left(\begin{array}{cc} -1 & 0 \\0 & 1 \end{array}\right) $$ (think of the symmetries of a square -- the matrix for $r$ is that of a rotation by $\pi/2$ with respect to any orthonormal basis, so pick one which is given by the eigenvectors of a reflection). It is then easy to see that $rs = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ (or geometrically, the two classes of reflections swap each others eigenlines) and so $\{1,r,s,rs\}$ is a basis of $\text{Mat}_2(\mathbb Q)$.
Of course you don't even really need to calculate anything after you see that the image of $D_8$ lies in $\text{Mat}_2(\mathbb Q)$, since the other option was a division algebra over $\mathbb Q$ of dimension $4$, but since everything can be worked out concretely with relatively little work, it seems better to be concrete.