jmc

We can substitute $x=y-1$ to obtain a polynomial having roots $a+1$, $b+1$, $c+1$, namely, $$(y-1{)}^3-(y-1)+1=y^3-3y^2+2y+1.$$The sum of the reciprocals of the roots of this polynomial is, by Vieta's formulas, $\frac{2}{-1}=\boxed{-2}$.