We have $\det M=(a+b)(c+d)$ where $M=\begin{bmatrix} a& 0& -1& 0\\ 0& c& 0& -1\\ b& 0& 1& 0\\ 0& d& 0& 1 \end{bmatrix}$ and $\det M'=(a'+b')(c'+d')$ 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 linears forms in $a,b,c,d,a',b',c',d',\pm1,0$ as entries that somehow uses $M$ and $M'$ as building blocks 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)?

It is ok to modify $M$ and $M'$ and use certain auxiliary supporting matrices as long as some structure of underlying matrices involved is evident.

Determinant of the following matrix suffices.

$$\begin{bmatrix} 1&0&0&0&0&0&0&0&c\\ 0&1&0&0&0&0&0&0&c\\ 0&0&1&0&0&0&0&0&d\\ 0&0&0&1&0&0&0&0&d\\ 0&0&0&0&1&0&0&0&c'\\ 0&0&0&0&0&1&0&0&c'\\ 0&0&0&0&0&0&1&0&d'\\ 0&0&0&0&0&0&0&1&d'\\ -a&-b&-a&-b&-a'&-b'&-a'&-b'&0 \end{bmatrix}$$

So we know such matrices exist. However there is a certain trouble here because it needs us to expand the polynomial fully and write down the determinant. There is no intuition of using $M$ and $M'$ as a building block.