Indeed, the answer is yes, as proved by Banks, Freiberg, and Turnage-Butterbaugh [BFT15] with an application of the Maynard-Tao machinery concerning bounded gaps between primes [Ma15]. They in fact prove that, for any $m\geq 1$, there are infinitely many $n$ such that \[d_n<d_{n+1}<\cdots <d_{n+m}\] and infinitely many $n$ such that \[d_n> d_{n+1}>\cdots >d_{n+m}.\]
Solved by Verstraëte [Ve05], who gave a non-constructive proof that such a set $A$ exists.
Liu and Montgomery [LiMo20] proved that in fact this is true when $A$ is the set of powers of $2$ (more generally any set of even numbers which doesn't grow too quickly) - in particular this contradicts the previous belief of Erdős.
This was solved by Beck [Be91], who proved that there exists some $c>0$ such that \[\max_{n\leq N} M_n > N^c.\]
This is true if $A$ is unbounded or dense in some interval. It therefore suffices to prove this when $A=\{a_1>a_2>\cdots\}$ is a countable strictly monotone sequence which converges to $0$.
Steinhaus [St20] has proved this is false whenever $A$ is a finite set.
This conjecture is known in many special cases (but, for example, it is is open when $A=\{1,1/2,1/4,\ldots\}$. For an overview of progress we recommend a nice survey by Svetic [Sv00] on this problem.