Eigenvalues of a linear transformation $M_{2,2}\ \rightarrow M_{2,2}$

Expanding my comment, it is clearer to write out what we want explicitly. We desire $\lambda$ and $A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\neq 0$ such that the following holds. $$g(\begin{bmatrix} a&b\\c&d \end{bmatrix}) = \begin{bmatrix} 0&d-a \\ a-d&0 \end{bmatrix}=\lambda\begin{bmatrix} a&b\\c&d \end{bmatrix}.$$

We separate for clarity the cases $\lambda=0$ and $\lambda\neq 0$.

For $\lambda=0$, we see that $d=a$ is the only restriction, so that $b,c$ can be chosen freely. Hence, the eigenspace corresponding to 0 is three dimensional. It has basis $$\begin{bmatrix} 0&1\\0&0 \end{bmatrix},\begin{bmatrix} 0&0\\1&0 \end{bmatrix},\begin{bmatrix} 1&0\\0&1 \end{bmatrix}.$$

For $\lambda\neq 0$, we see that $a=d=0$ since $\lambda a = \lambda d=0$. This forces $b=c=0$ since $\lambda c = a-d = 0 = d-a = \lambda b$. Thus there are no eigenvalues that are nonzero.

You are correct in your proof that $\lambda$ must be $0$.

However, you have then overlooked the eigenvector $\pmatrix{1\\0\\0\\1}$.