Here is one of many possible proofs.
Notice that, by the last equation, we can perform the same elementary row operations on B and BA simultaneously, and retain the equality.
Further note that row operations of the first kind: adding a multiple of one row to another, dont change the determinant, on either side of the equation.
We first notice that the claim is easy if the matrix B is diagonal (has all off diagonal elements equal to zero.)
For B diagonal we have (BA)_ij = (B_ii)(A_ij).
We can factor B_ii from each row in evaluating the determinant, getting BA= (_iB_ii)A; but the product of the diagonal elements is the determinant of a diagonal matrix, so that we get BA=BAin this special case.
If we can reduce B to diagonal form by elementary row operations of the first kind obtaining a diagonal matrix B, and perform the identical operations on BA to form (BA) we then have, as desired: BA=(BA)=BA=BA.
You can always reduce any matrix B to diagonal form by elementary row operations of the first kind unless B is singular and its determinant is 0. In that case the columns of B do not span the entire space, so that the columns of BA dont do so either, and BA must also be singular.