We have $\det M=(a+b)(c+d)$ where 
$M=\begin{bmatrix}
 a& 0& -1& 0\\
 0& c& 0& -1\\
 b& 0& 1& 0\\
 0& d& 0& 1
\end{bmatrix}$.

Is there a matrix $A$ with $a,b,c,d,a',b',c',d',\pm1,0$ as entries that gives
$$
\det A =
(a+b)(c+d)+(a'+b')(c'+d')
$$
(if needed we can use other $\Bbb Z$ entries but I would prefer not)?