Your question is quite interesting!
So, let's calculate: first, how can we write $\overline{a_1a_2}$? In fact, $a_1$ and $a_2$ are simply, as you said, "glued" together. So, we simply have $\overline{a_1a_2} = 10a_1+a_2$.
Now, let's re-write your equation using this result:
$$\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix} \times \begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pmatrix} = \begin{pmatrix}
\overline{a_1 a_2} & \overline{b_1 b_2} \\
\overline{c_1 c_2} & \overline{d_1 d_2}
\end{pmatrix} = \begin{pmatrix}
10a_1+a_2 & 10b_1+b_2 \\
10c_1+c_2 & 10d_1+d_2
\end{pmatrix}$$
Now, let's name this "equation" $(E)$ with every number being an integer.
Then, by calculating the product, we have
$$(E) \iff \begin{pmatrix} a_1a_2 + b_1c_2 & a_1b_2 + b_1d_2\\
c_1a_2 + d_1c_2 & c_1b_2 + d_1d_2 \end{pmatrix} = \begin{pmatrix}
10a_1+a_2 & 10b_1+b_2 \\
10c_1+c_2 & 10d_1+d_2
\end{pmatrix}$$
Finally, we are able to identify the coefficients and get a system of equations:
$$\boxed{(E) \iff \begin{cases} a_1a_2+b_1c_2 = 10a_1+a_2\\
c_1a_2+d_1c_2 = 10c_1+c_2 \\
a_1b_2+b_1d_2 = 10b_1+b_2\\
c_1b_2 + d_1d_2 = 10d_1+d_2 \end{cases}}$$
Now that we got this system, you can notice that there are only 4 equations for a total of 8 unknowns. So 4 of them will have to be fixed, then they'll give you the 4 other. Now it's up to you to find "groups" of integers that work!