Non-unique commutating cases of a determinant identity We have $$
\det \pmatrix{A&B\\ C&D}=
\begin{cases}
\det(AD-BC) & \text{ if } CD=DC,\\
\det(DA-CB) & \text{ if } AB=BA,\\
\det(DA-BC) & \text{ if } BD=DB,\\
\det(AD-CB) & \text{ if } AC=CA.
\end{cases}
$$
See for example http://www.ee.iisc.ac.in/new/people/faculty/prasantg/downloads/blocks.pdf.
What if two pairs $(A,C)$ and $(A,B)$ commute?
 A: Assume that all the matrices are $n\times n$.
Switch the column $j$ and $n+j$ for $1\leqslant j\leqslant n$ in order to be reduced to the case where the top and bottom right matrices commute. 
Do a similar operation to treat the case where the top left and right matrices commute.
A: If two pairs commute, say $AB=BA$ and $AC=CA$, then two of the identities hold: the big determinant is both equal to $\det(DA-CB)$, and equal to $\det(AD-CB)$. Feel free to use whichever one is convenient for your situation.
Perhaps your question is trying to find out: Assuming $AB=BA$ and $AC=CA$, is there a way to see directly that $\det(DA-CB)=\det(AD-CB)$? Here $A,B,C,D$ are $n \times n$ matrices ($D$ is quite arbitrary!).
First, assuming $A$ is invertible, we have
$$
  DA-CB = A^{-1}(AD-CB)A,
$$
using that $A^{-1}(CB)A = A^{-1}A(CB) = CB$, by the hypothesis. So $\det(DA-CB)=\det(A^{-1}(AD-CB)A) = \det(AD-CB)$.
Now, if $A$ is not invertible, it seems not quite so clear. Perhaps someone else will have a nice idea; off the top of my head, let's follow an idea described in [Silvester, 2000]. Introduce a variable $x$ and set $A_x = A + xI$ where $I$ is the $n \times n$ identity matrix. Then $A_x$ commutes with $B$ and $C$ because $A$ and $I$ do, so
$$
  A_x(D A_x - CB) = (A_x D - CB) A_x .
$$
Therefore $\det(A_x)\det(D A_x - CB) = \det(A_x D - CB) \det(A_x)$, identically as polynomials in $x$. Since $\det(A_x)$ is a monic polynomial, we must have $\det(D A_x - CB) = \det(A_x D - CB)$ identically. Setting $x=0$, so we have $A_0 = A+0I = A$, gives the desired equality.
