We have $a-b = \frac{1}{c} - \frac{1}{b} = \frac{b-c}{bc}$, $b-c = \frac{1}{a} - \frac{1}{c} = \frac{c-a}{ac}$, and $c-a = \frac{1}{b} - \frac{1}{a} = \frac{a-b}{ab}$. Multiplying these equations together yields $(a-b)(b-c)(c-a) = \frac{(b-c)(c-a)(a-b)}{(abc)^2}$. Since $a,b,c$ are pairwise distinct, $(a-b)(b-c)(c-a) \neq 0$, so $(abc)^2 = 1$, so $abc = \pm 1$. $(a,b,c) = (1, -\frac{1}{2}, -2)$ and $(a,b,c) = (-1, \frac{1}{2}, 2)$ yield solutions to the given equations satisfying $abc = 1$ and $abc = -1$, respectively, so the set of all possible values of $abc$ is $\{-1, 1\}$.