jmc

We solve for $b$ in the equation $ab=a-b.$ Adding $b$ to both sides gives $ab+b=a,$ or $b(a+1)=a.$ If $a=-1,$ then we have $b(0)=-1,$ which cannot be true, so we can safely divide by $a+1$ to get $$b=\frac{a}{a+1}.$$Then, substituting our expression for $b,$ we get $$\begin{array}{rl}\frac{a}{b}+\frac{b}{a}-ab& =\frac{a}{a/(a+1)}+\frac{a/(a+1)}{a}-a\cdot \frac{a}{a+1}\\ & =(a+1)+\frac{1}{a+1}-\frac{a^2}{a+1}\\ & =\frac{(a+1{)}^2+1-a^2}{a+1}\\ & =\frac{2a+2}{a+1}\\ & =\boxed{2},\end{array}$$which is the only possible value of the expression.