Note that we have
$$
\det(A + BC) = \det(A[I + A^{-1}BC]) = \det(A)\det(I + [A^{-1}B]C).
$$
with $U = A^{-1}B$ and $V = C$, it suffices to show that
$$
\det(I + UV) = 1 + VU.
$$
This can be seen as an instance of the W-A identity. One proof of this result is given in the linked page.
If you're still set on using the linked result, we can indeed prove this result as a consequence. Let $M$ denote the matrix
$$
M = \pmatrix{A & -B\\ C & 1}.
$$
If we apply the linked result directly to this matrix, then we have
$$
\det(M) = \det(A)\det(D - CA^{-1}(-B)) = \det(A)(1 + CA^{-1}B).
$$
On the other hand, we note that
$$
\det \pmatrix{A & -B\\ C & 1} =
(-1)^{n-1}\det\pmatrix{-B & A\\ 1 & C} =
[(-1)^{n-1}]^2\det\pmatrix{1 & C\\ -B & A}.
$$
That is, we have
$$
\det(M) = \det\pmatrix{1 & C\\ -B & A}.
$$
Applying the linked result to this matrix yields
$$
\det\pmatrix{1 & C\\ B & A} = \det(1)\det(A - (-B)1^{-1}C) = \det(A + BC).
$$
So, we have $\det(A)(1 + CA^{-1}B) = \det(A + BC)$ as desired.