I will add a few comments to Bjørn Kjos-Hanssen's answer.

<blockquote>
  <ol>
  <li>Does property (A) have a name in the literature? Is it a studied notion?</li>
  </ol>
</blockquote>

The property is a kind of independence property. A set-theorist's antichain $S$ of a lattice $L$ is called **weakly independent** if
whenever $a_1,\ldots,a_{n+1}\in S$ are distinct, then $(a_1\vee\cdots\vee a_n)\wedge a_{n+1} = 0$. It is called **strongly independent** if whenever $I$ and $J$ are finite sets of indices, then $(\bigvee_{i\in I} a_i)\wedge 
(\bigvee_{j\in I} a_j)=(\bigvee_{k\in I\cap J} a_k).$ Property (A) lies between these two concepts. In a modular lattice all three notions agree (weak independence, Property (A), and strong independence).

<blockquote>
  <ol start="2">
  <li>When does (A) coincide with being an antichain? Only in distributive lattices? </li>
  </ol>
</blockquote>

It coincides in any pseudocomplemented lattice. These need not be distributive. For example, the congruence lattice of a semilattice is pseudocomplemented and almost never distributive. The order-dual of the lattice of convex subsets of a finite chain is pseudocomplemented, but not distributive when the chain has more than 2 elements.

<blockquote>
  <ol start="3">
  <li>Does maximal with respect to (A) imply being a maximal antichain?</li>
  </ol>
</blockquote>

I would guess that being maximal with respect to (A) rarely means being a maximal antichain for lattices that are not pseudocomplemented. For a simple example, if $S$ is an antichain in the lattice $L$ of subspaces of a vector space $V$, and $S$ satisfies Property (A), then the size of $S$ cannot exceed the dimension of $V$. But $L$ can have antichains  larger than the dimension of $V$. For example, if $V$ has finite dimension greater than one and $T\subseteq L$ is the set of 1-dimensional subspaces, then $T$ is larger than any antichain in $L$ that has Property (A).