Suppose $D:\mathbb{R}_{2\times 2}\rightarrow \mathbb{R}$ is a function from the space of 2x2 matrices into the reals. Suppose also that it satisfies $D(AB)=D(A)D(B)$ for all matrices $A,B\in \mathbb{R}_{2\times 2}$. Call
$$J = \begin{bmatrix}0&1\\1&0\end{bmatrix}$$
and suppose $D(I)\ne D(J)$. I have already shown that such a function must map $D(I)=1$ and $D(J)=-1$, and $D(0)=0$. I've also shown that if $A^2=0$ then $D(A)=0$, and that if $A,B$ differ by a row-swap then $D(A)=-D(B)$.
However, I can't see my way through showing that if any matrix $A$ has a zero-row, then $D(A)=0$. I have considered squaring $A$, writing it as a product of two other matrices, multiplying it by other matrices, and using the fact that matrices don't commute but real products do. Nothing has worked and I'm out of ideas.
If $A$ has the first row $0$ iff $A = P \cdot A$, where $P=\begin{bmatrix}0&0\\0&1\end{bmatrix}$. Now it enough to show that $P$ maps to $0$. Consider also $Q= \begin{bmatrix}1&0\\0&0\end{bmatrix}$. We have $Q=JPJ^{-1}$, so $P$, $Q$ map to the same value. Moreover, we have $PQ=0$, the zero matrix, and so $PQ$ maps to $0$. We conclude $P$ maps to $0$.