Let $f(x) = x^3 - px^2 + qx - r$ be the monic polynomial with distinct roots $a, b, c$. From Vieta's, $a + b + c = p, ab + bc + ca = q,$ and $abc = r$. Since $\{a + b, b + c, c + a\} = \{ab, bc, ca\}$, we have $(a + b) + (b + c) + (c + a) = ab + bc + ca \iff 2p = q$, $(a + b)(b + c) + (b + c)(c + a) + (c + a)(a + b) = (ab)(bc) + (bc)(ca) + (ca)(ab) \iff (a + b + c)^2 + (ab + bc + ca) = abc(a + b + c)$ $\iff p^2 + q = pr \iff p(p - r + 2) = 0 \iff p = 0$ or $p = r - 2$, and $(a + b)(b + c)(c + a) = (ab)(bc)(ca) \iff (a + b + c)(ab + bc + ca) - abc = (abc)^2 \iff pq - r = r^2 \iff 2p^2 = r^2 + r$. If $p = 0$, then $q =0$ and $r^2 + r = 0 \iff r = 0$ or $r = -1$. So $f(x) = x^3$ or $x^3 + 1$. But both of them do not have three distinct roots, contradiction. If $p = r - 2$, then $2(r - 2)^2 = r^2 + r \iff r^2 - 9r + 8 = 0 \iff r = 1$ or $r = 8$. If $r = 8$, then $p = 6$ and $q = 12$, giving us $f(x) = x^3 - 6x^2 + 12x - 8 = (x - 2)^3$, which does not have three distinct roots. So the only possible case is when $r = 1$ for which $p = -1, q = -2$, and $f(x) = x^3 + x^2 - 2x - 1$. Now we will show that $f(a^2 - 2) = f(b^2 - 2) = f(c^2 - 2) = 0$. Notice that $f(x^2 - 2) = (x^2 - 2)^3 + (x^2 - 2)^2 - 2(x^2 - 2) - 1 = x^6 - 5x^4 + 6x^2 - 1$ $= (x^3 - 2x)^2 - (x^2 - 1)^2 =(x^3 + x^2 - 2x - 1)(x^3 - x^2 - 2x + 1) = f(x)(x^3 - x^2 - 2x + 1)$. Thus, since $f(a) = f(b) = f(c) = 0$, we have $f(a^2 - 2) = f(b^2 - 2) = f(c^2 - 2) = 0$. Therefore, $a^2 - 2, b^2 - 2, c^2 - 2$ are the roots of $f(x)$ and $\{a, b, c\} = \{a^2 - 2, b^2 - 2, c^2 - 2\}$.