 Solution

The same matrix similarity preserves being a diagonalisable matrix. If $S^{-1}(AB)S=D$ is a diagonal matrix, thenTherefore $BA$ is a diagonalisable matrix whenever $AB$ is, and the converse follows by interchanging $A,B$. Thus they are either both diagonalisable matrices or neither is. In fact the previous part shows that one invertible factor suffices.