When is a bounded lattice isomorphic to some $(\mathbf{2}^S,\subseteq)$? If I have a set $S$, I can turn it into a bounded lattice in any number of ways, the simplest of which is to designate an initial object and a terminal object, adding a single object if necessary. Such an extension is clearly not isomorphic to a power set unless $|S|\leq 4$.
On the other hand, if I have a bounded lattice $L$, is there some property strictly in terms of $\lor$, $\land$, and $\leq$ that guarantees that $L$ is isomorphic to a power set? In other words, is there a set of properties of $L$ implying $\exists S_L$ s.t. $L\simeq(\mathbf{2}^{S_L},\subseteq)$ without explicitly constructing $L\leftrightarrow(\mathbf{2}^{S_L},\subseteq)$ as a part of the proof for each specific $L$?

As a starting point, it seems like if $\exists S$ s.t. $L\simeq(\mathbf{2}^S,\subseteq)$ then we can define $S$ as the set of completely join prime objects of $L$. For fixed $x\in L$, we can then define the subset $S_x\equiv\{y\in S\mid y\leq x\}\in\mathbf{2}^S$.
Additionally, I'm fairly certain that $L$ must be atomic and that the atoms of $L$ must be exactly $S\setminus$, i.e., that $L$ can be expressed strictly in terms of joins on subsets of $S$.
The problem with this approach is that it requires showing that each object in $L$ can be constructed from $S$ and that $\mathbf{2}^S$ is fully represented in $L$. It would be nice if there was some existing proof for simple conditions on $L$ that guarantees that such an $S$ can be found without needing to find it.

(To put the question in context, I'm interested in a specific situation where I want to extend a posetal category by including all products and coproducts. I'd like to know what the uniqueness requirements are for this to be isomorphic to a power set.)
 A: One such characterization is that a poset $L$ is isomorphic to a power set iff $L$ is a complete lattice (every subset has a join and a meet), complemented (for each $x\in L$ there exists $\neg x\in L$ such that $x\wedge \neg x=0$ and $x\vee \neg x=1$), and completely distributive, meaning it satisfies the infinite distributive law $$\bigwedge_{i\in I}\bigvee_{j\in J}a_{ij}=\bigvee_{f:I\to J}\bigwedge_{i\in I}a_{if(i)}$$ for arbitrary index sets $I$ and $J$.
This characterization is well-known, and you can probably find it in any text that covers the basic theory of complete Boolean algebras (a Boolean algebra is a complemented distributive lattice, so the characterization above says $L$ is isomorphic to a power set iff it is a complete Boolean algebra that is completely distributive).  As a sketch of a proof, given such an $L$, note that by complete distributivity $$1=\bigwedge_{a\in A}(a\vee \neg a)$$ is equal to the join of all elements of the form $\bigwedge_{a\in A}a^*$, where each $a^*$ is either $a$ or $\neg a$.  But if such an element $b=\bigwedge_{a\in A}a^*$ is nonzero, then it is an atom, since if $0<c<b$ then $b\not\leq c$ and also $b\not\leq \neg c$ which is impossible since either $c$ or $\neg c$ is a term in the meet defining $b$.  So, $1$ is a join of atoms in $L$, and it follows from distributivity that every element of $L$ is the join of the atoms below it.  Now let $S$ be the set of atoms of $L$, let $f:L\to \mathbf{2}^S$ send each element of $L$ to the set of atoms below it, and let $g:\mathbf{2}^S\to L$ send each set of atoms to its join.  It is clear that $f$ and $g$ are both order-preserving, and we know $gf=1_L$.  Distributivity can be used to show that $f$ preserves arbitrary joins, and so $f$ is surjective since $L$ is complete and the image of $f$ contains all singletons.  So, $f$ is a bijection and $g=f^{-1}$, so $f$ is an order-isomorphism.
