jmc

From the equation $a^3+b^3=a+b,$ $$(a+b)(a^2-ab+b^2)=a+b.$$Since $a$ and $b$ are positive, $a+b$ is positive, so we can cancel the factors of $a+b$ to get $$a^2-ab+b^2=1.$$Then $$\frac{a^2+b^2-1}{ab}=\frac{ab}{ab}=\boxed{1}.$$