Let $G = D_8$ be the dihedral group of order $8$, i.e., \begin{align*} G = \langle a,b \mid a^4 = 1 = b^2, ab = ba^{-1} \rangle. \end{align*} By standard results from ring/represention theory, the group algebra $\mathbb{C} G$ decomposes as a product of matrix rings as follows: \begin{align*} \mathbb{C}G \cong \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}). \end{align*} I would like to explicitly find elements of $\mathbb{C} G$ which behave like the elements $(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,e_{ij})$, where $e_ij$ is the matrix with a $1$ in the $(i,j)$th position, and zeros elsewhere.
I've found the following pairwise orthogonal idempotents which hopefully correspond to $1$ in each copy of $\mathbb{C}$:
\begin{align*} e_1 = \tfrac{1}{8}(1+a+a^2+a^3+b+ab+a^2b+a^3b)\\ e_2 = \tfrac{1}{8}(1+a+a^2+a^3-b-ab-a^2b-a^3b)\\ e_3 = \tfrac{1}{8}(1-a+a^2-a^3+b-ab+a^2b-a^3b)\\ e_4 = \tfrac{1}{8}(1-a+a^2-a^3-b+ab-a^2b+a^3b) \end{align*}
The element \begin{align*} e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3+b+iab-a^2b-ia^3b)\\ \end{align*}
is also an idempotent and is orthogonal to the $e_i$, and so it makes sense to set $e_{22} = 1-e_1-e_2-e_3-e_4-e_{11}$. However, I can't seem to find suitable elements for $e_{12}$ and $e_{21}$. Am I going in the right direction at the moment?
(Also, my method has been largely ad hoc; is there a uniform method for approaching this sort of problem for an arbitrary finite group $G$?)
I managed to work this out, so I'll provide my solution, which carries on from my work in the question.
The elements \begin{align*} e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3) \\ e_{22} = \tfrac{1}{4}(1-ia-a^2+ia^3) \end{align*} are orthogonal idempotents which are also orthogonal to each of the $e_i$. Observe that $e_{11}b = be_{22}$ and $be_{11} = e_{22}b$. These identities ensure that if we set \begin{align*} e_{12} = e_{11}b = \tfrac{1}{4}(b+iab-a^2b-ia^3b) \\ e_{21} = e_{22}b = \tfrac{1}{4}(b-iab-a^2b+ia^3b), \end{align*} then the correct relations hold between these matrices. Additionally, one can check that $e_1+e_2+e_3+e_4+e_{11}+e_{22} = 1$. Finally, the matrix \begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 & i & -i & 0 & 0 \\ 1 & 1 & 1 & 1 & -1 & -1 & 0 & 0 \\ 1 & 1 & -1 & -1 & -i & i & 0 & 0 \\ 1 & -1 & 1 & -1 & 0 & 0 & 1 & 1 \\ 1 & -1 & -1 & 1 & 0 & 0 & i & -i \\ 1 & -1 & 1 & -1 & 0 & 0 & -1 & -1 \\ 1 & -1 & -1 & 1 & 0 & 0 & -i & i \end{pmatrix} \end{align*} has nonzero determinant, so the eight elements we have found are linearly independent. It follows that the map \begin{gather*} \phi: \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}) \to \mathbb{C}D_8 \\ e_1 \mapsto \tfrac{1}{8}(1+a+a^2+a^3+b+ab+a^2b+a^3b)\\ e_2 \mapsto \tfrac{1}{8}(1+a+a^2+a^3-b-ab-a^2b-a^3b)\\ e_3 \mapsto \tfrac{1}{8}(1-a+a^2-a^3+b-ab+a^2b-a^3b)\\ e_4 \mapsto \tfrac{1}{8}(1-a+a^2-a^3-b+ab-a^2b+a^3b)\\ e_{11} \mapsto \tfrac{1}{4}(1+ia-a^2-ia^3) \\ e_{22} \mapsto \tfrac{1}{4}(1-ia-a^2+ia^3) \\ e_{12} \mapsto \tfrac{1}{4}(b+iab-a^2b-ia^3b) \\ e_{21} \mapsto \tfrac{1}{4}(b-iab-a^2b+ia^3b) \end{gather*} is a surjective ring homomorphism between $8$-dimensional spaces, and hence an isomorphism.