jmc

First we find that $f(g(x))=A(Bx+A)+B=ABx+A^2+B$ and $g(f(x))=B(Ax+B)+A=ABx+B^2+A$.

Now we plug back in. $$\begin{array}{rl}f(g(x))-g(f(x))& =B-A\\ (ABx+A^2+B)-(ABx+B^2+A)& =B-A\\ A^2-B^2+B-A& =B-A\\ A^2-B^2& =0\\ (A-B)(A+B)& =0\end{array}$$

Since we are given that $A\ne B$, this means that $A+B=\boxed{0}.$