Your work so far is fine (watch the capital P on the LHS, though). On the RHS, you can get a common denominator and add. The LHS factors, and then you can solve for L(y). What do you get?
Oh, dear, oh dear, oh, dear! I really dislike the Laplace Transform! It can be used in two diametrically opposite ways- to "look up" solutions to non-homogeneous equations or to assert a form of a solution in very abstract theorems. The first is more easily done in other ways and the second, undergraduates never see.
I would look at that equation and immediately write down it characteristic equation: . It's easy to see, by inspection, that r= 1 is a root of that and that so that r= 1 is a double root and r= -1 is a root. That tells us that the general solution to the associated homogeneous equation is .
The "non-homogeneos" part is . Since is already a solution to the associated homogeneous equation, we try a solution of the form . Then , , and .
Putting those into the differential equation will give you equations for A and B.