The answer is "no". Consider the ring of matrices with first row $(0, x,y,...,z)$ and all other entries 0. This is a ring with zero product, so commutative. It is not isomorphic to a subring of $D$.
Two caveats, and some high-level hints:
First, the claim you're asked to prove is not necessarily true! There exist proper subsets of $\mathbb C$ that are field-isomorphic to $\mathbb C$ itself, and such a set will map to a proper subset of $A$ that is isomorphic to $\mathbb C$ but certainly doesn't have the shape specified for $B$ in the problem.
(To prove that such a subset exists requires the axiom of choice, but it won't help us to decide to work in ZF -- if we prove the claim from ZF we would have disproved the axiom of choice, and a result of Gödel (no, not the famous one) states that the axiom of choice cannot be disproved in ZF -- unless ZF is inconsistent, in which case they sky is falling anyway).
In order to fix this and get something that can be proved, we need to revise the claim in one of two ways:
Replace "isomorphic to $\mathbb C$ as a field" with "isomorphic to $\mathbb C$ as an $\mathbb R$-algebra" -- that is, require that scaling a matrix in $B$ by a real constant corresponds, via the isomorphism, to multiplying the complex number by that same real number. Or,
Require that $B$ is a closed subset of $\mathbb R^{2\times 2}$.
Second, it is not really clear just what was meant by "isomorphic to $\mathbb C$ as a field" in the first place. For example, if $D$ is the set of $\mathbb 3\times 3$ matrices of the form
$$ \begin{bmatrix}a & -b & 0 \\ b & a & 0 \\ 0 & 0 & 0 \end{bmatrix} $$
should we then consider $D$ to be "isomorphic to $\mathbb C$ as a field"? It definitely is isomorphic to $\mathbb C$ as an $\mathbb R$-algebra, but $1\in\mathbb C$ is not represented by the identity element of $\mathbb R^{3\times 3}$. Now, the concept of a field homomorphism does require that $1$ maps to $1$, but $\mathbb R^{3\times 3}$ is not a field, and it is not a priori clear whether or not viewing a subset of it as a field should require us to use the particular identity element from the surrounding non-field.
Back in the $2\times 2$ case: We assume that an isomorphism $\varphi:\mathbb C\to B$ has been given. If we can't assume that $\varphi(1) = I_{2\times 2}$, then the solution will need some additional footwork to get started. In particular, then we don't automatically know that $\varphi(i)$ must be a solution of $X^2 = -I_{2\times 2}$. We can in fact still get through by instead starting with $X^3 = -X$ (together with $X\ne 0$), but there are additional details to keep track of in that case.
In each of these cases, some thinking about either $X^2 = -I$ or $X^3 = -X$ should enable you to find an invertible basis change $g$ such that
$$X = g\begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}g^{-1}.$$
If you don't yet know that $\varphi(1) = I_{2\times 2}$, computing $-X^2$ in terms of $g$ will show it now.
If you have decided that $B$ must be an $\mathbb R$-algebra and isomorphic to $\mathbb C$ as such, it is now easy to conclude that $B$ is in fact identical to $gAg^{-1}$.
However, if all you know is that $B$ is closed, it's not quite that straightforward. Now that you know that $I\in B$, you can show that $\lambda I \in B$ for every real $\lambda$ -- but beware that this doesn't guarantee that $\lambda I=\varphi(\lambda)$! Since $B$ is closed under matrix multiplication and addition, we can then at least conclude $gAg^{-1} \subseteq B$. But an additional argument will be needed to show that $B$ cannot be larger than $gAg^{-1}$. One strategy could be to show that if there is a $b\in B$ with $g^{-1}bg\notin A$, then you can find an nonzero non-invertible element in $B$ -- which certainly cannot represent anything from $\mathbb C$.
Best Answer
Are you familiar with $\Bbb{R}[\omega]=\{a+b\omega \, | \, a,b \in \Bbb{R}\}$, Eisenstein ring, where $\omega$ is the cube root of unity?
Here the norm $|a+b\omega|$ is defined as $(a+b\omega)(a+b\omega^2)=a^2-ab+b^2$.
If you are "determined" you can find the isomorphism:-)