jmc

The definition of $f$ lets us evaluate $f(2)$: $$f(2)=\frac{b}{2\cdot 2-3}=\frac{b}{1}=b.$$Therefore we want to find all possible $b$ for which $$b=f^{-1}(b+1).$$This is equivalent to $$f(b)=b+1.$$When we substitute $x=b$ into the definition of $f$ we get $$f(b)=\frac{b}{2b-3},$$so we are looking for all solutions $b$ to the equation $$\frac{b}{2b-3}=b+1.$$Assuming $b\ne \frac{3}{2}$, we can multiply both sides by $2b-3$ to get $$b=(2b-3)(b+1)=2b^2-b-3,$$so $2b^2-2b-3=0$. We note that $b=\frac{3}{2}$ is not a solution. By Vieta's formulas, the product of the roots of the quadratic equation $a{x}^2+bx+c=0$ is $c/a$, so in this case, the product of the roots is $\boxed{-\frac{3}{2}}$.