det(I + AB) = det(I + BA) by density

Idea. An identity between two continuous functions that holds on a dense set holds everywhere. The invertible matrices are dense, and for them the identity is a one-line conjugation.

Step 1: $A$ invertible. Then
$$A(I+BA)A^{-1}=A A^{-1}+ABAA^{-1}=I+AB.$$
So $I+AB$ and $I+BA$ are conjugate, hence have the same determinant (because $\det(PXP^{-1})=\det X$).

Step 2: density. Fix $B$. The two functions
$$f(A)=\det(I+AB),\qquad g(A)=\det(I+BA)$$
are polynomials in the entries of $A$ (a determinant is a polynomial in the entries, and the entries of $I+AB$ are polynomials in those of $A$). In particular they are continuous on $M_n(\mathbb{R})$.

For an arbitrary $A$, the matrices $A_\varepsilon=A+\varepsilon I$ are invertible for all but finitely many $\varepsilon$ (their determinant is a nonzero polynomial in $\varepsilon$ of degree $n$, equal to $\chi_{-A}(\varepsilon)$ up to sign). Hence $f(A_\varepsilon)=g(A_\varepsilon)$ for small $\varepsilon\ne0$, and letting $\varepsilon\to0$ gives $f(A)=g(A)$.

Remark. The same proof gives the more general identity $\det(\lambda I+AB)=\det(\lambda I+BA)$, and later the equality of the characteristic polynomials of $AB$ and $BA$. Note that it also holds for rectangular matrices $A\in M_{n,m}$, $B\in M_{m,n}$ in the form $\det(I_n+AB)=\det(I_m+BA)$, but then the conjugation argument must be replaced by a block-matrix computation.

Lesson. To prove an identity between polynomial functions of a matrix, prove it for invertible (or diagonalisable) matrices and conclude by density.