Let $A_1$ be the matrix algebra consisting of matrices of the form $$ \pmatrix{ \alpha & 0 \\ 0 & \beta\\ }$$ and let $A_2$ be the matrix algebra consisting of matrices of the form $$ \pmatrix{ \alpha & \beta \\ 0 & \alpha\\ }.$$ I want to prove that
these algebras are not isomorphic and that every two-dimensional unital complex Banach algebra is isomorphic to one of them.
I have proved the first part:
If such isomorphism exists, it would have to be multiplicative and bijective. Hence, a idempotent matrix in $A_1$ would be transformed to idempotent matrix in $A_2$. But those algebras have different numbers of idempotent algebras; a contradiction.
Sadly, I have been defeated by the second part. Anyone can help?
Let $B$ be a two-dimensional Banach algebra. Pick $x \in B \setminus \mathbb{C}\cdot e$. Since $\{e,x\}$ is a basis of $B$, we have
$$x^2 = c\cdot e + d\cdot x$$
for some $c,d\in \mathbb{C}$. Now let $\lambda\in\mathbb{C}$ a zero of $z^2 - d\cdot z - c = 0$, then
$$(x-\lambda\cdot e)^2 = x^2 - 2\lambda x + \lambda^2\cdot e = (d-2\lambda)x + (c+\lambda^2)e = (d-2\lambda)(x-\lambda\cdot e) + (c+\lambda d - \lambda^2)e,$$
so setting $y = x-\lambda e$ we have $y^2 = (d-2\lambda)y$ and $y\neq 0$.
If $d - 2\lambda = 0$, i.e. $y^2 = 0$, consider the map
$$(\alpha\cdot e + \beta\cdot y) \mapsto \begin{pmatrix}\alpha &\beta\\ 0 &\alpha\end{pmatrix}.$$
If $d-2\lambda\neq 0$, set $p = \frac{1}{d-2\lambda}y$ to obtain an idempotent $p \notin \{0,1\}$, let $q = e-p$. Then $\{p,q\}$ is a basis of $B$, and you can consider the map
$$(\alpha p + \beta q) \mapsto \begin{pmatrix}\alpha&0\\0&\beta \end{pmatrix}.$$