That should be doable by combining
(1) the Chinese Remainder Theorem to reduce to $m$ being a prime power (namely the Kloosterman sum factors as a product of Kloosterman sums module the primes powers dividing $m$; the other arguments do change, so this is called "twisted multiplicativity" in the literature).
(2) For $m=p^k$ a prime power, there are two cases:
(2.1) it is known (looking at congruences modulo the ideal $(1-e(1/p))$ of the $p$-th cyclotomic field) that the Kloosterman sum is non-zero if $k=1$;
(2.2) if $k\geq 2$, there is an exact formula in terms of the roots of the equation $x^2=ab$ modulo $p$, at least if $p$ is odd (I don't know what happens for $p=2$, but in principle it should be something similar). For instance, see Exercise 1 in Chapter 12 of Iwaniec-Kowalski. This formula shows exactly which Kloosterman sums vanish in that case.
It seems that for the specific question, the only tricky part is to handle what happens for the prime $2$.