If the determinants are 0, the theorem is obviously false, but what happens otherwise? And if it does not hold, could we add some other conditions to make it true?
My motivation for this came to me while trying to prove the elementary fact that if $ \lambda_1, ..., \lambda_n$ are eigenvalues of a matrix A, then $ detA = \lambda_1...\lambda_n$. Now I'm aware of the quick proof for this using the characteristic polynomial evaluated at 0, but I found my attempt more interesting.
First, I used the fact that A can be transformed into a reduced form $ \pmatrix{c_1 & . & . \\ . & .... & . \\ 0 & ... & 0 & c_n }$ with $ \det{A} = c_1...c_n$. To prove the theorem, we'd obviously want to show that $ c_1, ..., c_n$ are precisely the roots of the characteristic polynomial for A, that is $ \det{(A - \lambda I)}$. But they are precisely the roots of the polynomial $ (c_1-\lambda)...(c_n-\lambda) = \det{\pmatrix{c_1 - \lambda & . & . \\ . & .... & . \\ 0 & ... & 0 & c_n - \lambda}}$. So, we should also get that the last equation equals $ \det{(A-\lambda I)}$. So if I could somehow show that $ A-\lambda I $ can be transformed by gaussian elimination into $ \pmatrix{c_1 - \lambda & . & . \\ . & .... & . \\ 0 & ... & 0 & c_n - \lambda} $, my proof would work. And that's what motivated the question.
Any invertible matrix can be row-reduced to the identity matrix and back, which means that you can row-reduce any invertible matrix into any invertible matrix (of the same size, obviously).
Now, that said, row-reduction does not preserve determinants, so I fail to see how your argument goes.