No, neither implication holds.
If you take $\mathcal L$ to be the 8-element Boolean lattice, then remove one atom, the resulting poset is a selftop adiamond lattice which fails your condition. That is, the three coatoms meet to $0$, but one subset of two coatoms also meets to $0$.
On the other hand, start with any nonempty finite lattice $\mathcal K$ and form the poset that is the parallel sum of $\mathcal K$ and a singleton $z$. Now adjoin a top element $1$ and a bottom element $0$ to this poset. This is a selftop lattice with two maximal elements ($z$ and the top of $\mathcal K$), and no one of the coatoms is $0$. Hence this lattice satisfies your condition that no proper subset of coatoms meets to $0$. But there is no reason for this lattice to be adiamond, since $\mathcal K$ is an interval of arbitrary structure.