I'll come at this from a slightly different perspective:
The only two derangements of a set of three digits are their cyclic permutations. This means that two numbers in question are $100 A + 10 B + C$ and $100 B + 10 C + A$. We therefore have $$101A + 110 B + 11 C = 1000,$$ which we can rearrange to $$(100A + 110 B + 10C) + A + C = 1000.$$
Since the right-hand side is divisible by 10, and the terms in brackets on the left-hand side are automatically divisible by 10, we have must have $A + C$ divisible by 10 as well; and since they are single digits, we conclude that $A + C = 10$ or $A = C = 0$. In the latter case, we have $110B = 1000$, or $11B = 100$; but 100 is not divisible by 11, so this is impossible. Thus, $A + C = 10$.
Substituting this in, we obtain $101 A + 110 B + 11(10 - A) = 1000$, which reduces to $$9A + 11 B = 89$$
Since 9 and 11 are relatively prime, a solution can be obtained by repeatedly subtracting 9 from 89 until we get a multiple of 11; we have $89 - 5(9) = 44$, and so a solution is $A = 5$, $B = 4$. All other solutions are then of the form $A = 5 + 11 m$, $B = 4 - 9 m$ for some integer $m$; but none of these solutions are single-digit numbers.
Thus, the only solution is $$A = C = 5, \quad B = 4$$