jmc

Since $f(f^{-1}(x))=x$, it follows that $a(bx+a)+b=x$, which implies $abx+a^2+b=x$. This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$.

Then $b=-a^2$. Substituting into the equation $ab=1$, we get $-a^3=1$. Then $a=-1$, so $b=-1$, and $$f(x)=-x-1.$$Likewise $$f^{-1}(x)=-x-1.$$These are inverses to one another since $$f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.$$$$f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.$$Therefore $a+b=\boxed{-2}$.