If one side of the rectangle is not divisible by a, and the other is not divisible by b, then the tiling is impossible. Thus the number of bxb starting at row 2 must BE an a-multiple. Every square tile covers an a-multiple of black squares.
Every square tile covers an a-multiple of black squares.
This statement of how to position the colouring isn't good enough, I'm afraid. Suppose N columns are black with this positioning. The difference between these two numbers of black cells is H, which is not a multiple of a.