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)?