There is an isomorphism $f:\mathbb{C\times C}\rightarrow (M_2(\mathbb{R}),+,*)$ given by

$$f((1,0))=\left(\begin{array}{cc}
-1 & -1\\-1 & -1
\end{array}\right)\!/2,\ \ \
f((i,0))=\left(\begin{array}{cc}
1 & -1\\1 & -1
\end{array}\right)\!/2\sqrt{3},
$$

$$f((0,1))=\left(\begin{array}{cc}
-1 & -1\\1 & 1
\end{array}\right)\!/2,\ \ \
f((0,i))=\left(\begin{array}{cc}
1 & -1\\-1 & 1
\end{array}\right)\!/2\sqrt{3}
$$

which I found by finding the identity for *, then finding two idempotents which add up to the identity, then finding two elements whose squares are the negatives of those idempotents.

So there is an isomorphism
$g:(M_2(\mathbb{Z}),+,*) \rightarrow \mathbb{Z[\omega]\times Z[\omega]}$ given by

$$g\left(\left(\begin{array}{cc}
1 & 0\\0 & 0
\end{array}\right)\right)
= \left(\omega,\ \omega\right)
,\ \ \
g\left(\left(\begin{array}{cc}
0 & 1\\0 & 0
\end{array}\right)\right)
= \left(\omega^2,\omega^2\right)
,\\
g\left(\left(\begin{array}{cc}
0 & 0\\1 & 0
\end{array}\right)\right)
= \left(\omega, -\omega\right)
,\ \ \
g\left(\left(\begin{array}{cc}
0 & 0\\0 & 1
\end{array}\right)\right)
= \left(\omega^2,-\omega^2\right)
$$
Here $\omega=(-1+\sqrt{3i})/2$, and there are copies of $\mathbb{Z}$ in the image because $1=-(\omega+\omega^2)$.

youto find a structure, shouldn't you be trying to do this rather than asking other people to do it? Or have I misunderstood your meaning? $\endgroup$ – Yemon Choi Oct 28 '18 at 0:34