Prove $\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A-B)\det(A+B)$, even when $A$ and $B$ are not commutative. I am aware of the following identity
$\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B)$
When $A = D$ and $B = C$ and when $AB = BA$ the above identity becomes
$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = \det(A)\det(A - BA^{-1}B) = \det(A^2 - B^2) = \det(A-B)\det(A+B)$.
However, I couldn't prove this identity for the case where $AB \neq BA$.
EDIT: Based on @Trebor 's suggestion.
I think I could do the following.
$\det\begin{bmatrix}A & B\\ B & A\end{bmatrix} = 
\det\begin{bmatrix}A & B\\ B-A & A-B\end{bmatrix} = \det(A^2-B^2) = \det(A-B)\det(A+B)$.
 A: $\begin{pmatrix} A & B \\ B & A \end{pmatrix}\xrightarrow{\text{row1 -= row2}}\begin{pmatrix} A-B & B-A \\ B & A\end{pmatrix}\xrightarrow{\text{col2 += col1}}\begin{pmatrix} A-B & O  \\ B & A+B\end{pmatrix}$
A: Let's say $A, B$ are $n \times n$ matrices with entries from a field with characteristic $\ne 2{}^{\color{blue}{[1]}}$.
Let $I$ be the $n \times n$ identity matrix and $J = \begin{bmatrix}I & I \\ -I & I\end{bmatrix}$. Since $\det J = 2^n \ne 0$, $J$ is invertible.
Notice $$
J \begin{bmatrix}A & B \\ B & A\end{bmatrix}
= \begin{bmatrix}A+B & A+B \\ B-A & A-B\end{bmatrix}
= \begin{bmatrix}A+B & 0 \\ 0 & A-B\end{bmatrix} J
$$
We have
$$\det\begin{bmatrix}A & B \\ B & A\end{bmatrix}
= \det\begin{bmatrix}A+B & 0 \\ 0 & A-B\end{bmatrix}
= \det(A+B)\det(A-B)\tag{*1}
$$
Notes

*

*$\color{blue}{[1]}$ - As demonstrated by @Just a user's answer,
the requirement that entries from a field with characteristic $\ne 2$ can be dropped. $(*1)$ continues to work when entries of $A,B$ take values from any commutative ring.
Aside from using row/column operations as in @Just a user's answer, we can use the fact that LHS and RHS of $(*1)$ are polynomials with integer coefficients in entries of $A,B$. Since they are equal as a polynomial, it remains equal when we substitute the entries by elements from any commutative ring.
A: Your second attempt almost works. You may use the fact that $\det\pmatrix{A&B\\ C&D}=\det(AD-BC)$ whenever $A,B,C,D$ are square matrices of the same sizes and $CD=DC$:
$$
\det\pmatrix{A&B\\ B&A}
=\det\pmatrix{A&B\\ B-A&A-B}
=\det\left(A(A-B)-B(B-A)\right)
=\det\left((A+B)(A-B)\right).
$$
