$\let\bez\smallsetminus$ Joel’s answer resolves the question as stated, but I feel it should be pointed out that cofinal subsets of Boolean algebras without top have a simple characterization:

• If a poset $S$ can be embedded as a cofinal subset of $B\bez\{1\}$ for some Boolean algebra $B$, then $S$ is dually separative: for every $x,y\in S$ such that $x\nleq y$, there exists $z\ge y$ such that $x$ and $z$ have no common upper bound in $S$.
  [Take any $z\ge \neg x\lor y$.]

• If $S$ is a dually separative poset, there exists an embedding of $S$ into a complete Boolean algebra $B$ which maps $S$ onto a cofinal subset of $B\bez\{1\}$, and preserves all existing suprema and infima in $S$. (In particular, if $S$ is a meet semilattice, it will be a semilattice embedding.)
  [Make $S$ a topological space by declaring downwards closed subsets to be closed, let $B$ be the algebra of regular closed subsets of $S$, and embed $S$ into it by mapping $x$ to $\{y\in S:y\le x\}$.]

For $B$ to be atomless, it is necessary and sufficient that $S$ have no maximal elements.

(I hope I dualized it correctly, these results are usually stated for downwards cofinal subsets of $B\bez\{0\}$.)

$\let\bez\smallsetminus$ Joel’s answer resolves the question as stated, but I feel it should be pointed out that cofinal subsets of Boolean algebras without top have a simple characterization:

• If a poset $S$ can be embedded as a cofinal subset of $B\bez\{1\}$ for some Boolean algebra $B$, then $S$ is dually separative: for every $x,y\in S$ such that $x\nleq y$, there exists $z\ge y$ such that $x$ and $z$ have no common upper bound in $S$.
  [Take any $z\ge \neg x\lor y$.]

• If $S$ is a dually separative poset, there exists an embedding of $S$ into a complete Boolean algebra $B$ which maps $S$ onto a cofinal subset of $B\bez\{1\}$, and preserves all existing suprema and infima in $S$. (In particular, if $S$ is a meet semilattice, it will be a semilattice embedding.)
  [Make $S$ a topological space by declaring downwards closed subsets to be closed, let $B$ be the algebra of regular closed subsets of $S$, and embed $S$ into it by mapping $x$ to $\{y\in S:x\nleq y\}$.]

For $B$ to be atomless, it is necessary and sufficient that $S$ have no maximal elements.

(I hope I dualized it correctly, these results are usually stated for downwards cofinal subsets of $B\bez\{0\}$.)

$\let\bez\smallsetminus$ Joel’s answer resolves the question as stated, but I feel it should be pointed out that cofinal subsets of Boolean algebras without top have a simple characterization:

• If a poset $S$ can be embedded as a cofinal subset of $B\bez\{1\}$ for some Boolean algebra $B$, then $S$ is dually separative: for every $x,y\in S$ such that $x\nleq y$, there exists $z\ge y$ such that $x$ and $z$ have no common upper bound in $S$.
  [Take any $z\ge \neg x\lor y$.]

• If $S$ is a dually separative poset, there exists an embedding of $S$ into a complete Boolean algebra $B$ which maps $S$ to a cofinal subset of $B\bez\{1\}$, and preserves all existing suprema and infima in $S$. (In particular, if $S$ is a meet semilattice, it will be a semilattice embedding.)
  [Make $S$ a topological space by declaring downwards closed subsets to be closed, let $B$ be the algebra of regular closed subsets of $S$, and embed $S$ into it by mapping $x$ to $\{y\in S:y\le x\}$.]

(I hope I dualized it correctly, these results are usually stated for downwards cofinal subsets of $B\bez\{0\}$.)

$\let\bez\smallsetminus$ Joel’s answer resolves the question as stated, but I feel it should be pointed out that cofinal subsets of Boolean algebras without top have a simple characterization:

• If a poset $S$ can be embedded as a cofinal subset of $B\bez\{1\}$ for some Boolean algebra $B$, then $S$ is dually separative: for every $x,y\in S$ such that $x\nleq y$, there exists $z\ge y$ such that $x$ and $z$ have no common upper bound in $S$.
  [Take any $z\ge \neg x\lor y$.]

• If $S$ is a dually separative poset, there exists an embedding of $S$ into a complete Boolean algebra $B$ which maps $S$ onto a cofinal subset of $B\bez\{1\}$, and preserves all existing suprema and infima in $S$. (In particular, if $S$ is a meet semilattice, it will be a semilattice embedding.)
  [Make $S$ a topological space by declaring downwards closed subsets to be closed, let $B$ be the algebra of regular closed subsets of $S$, and embed $S$ into it by mapping $x$ to $\{y\in S:y\le x\}$.]

For $B$ to be atomless, it is necessary and sufficient that $S$ have no maximal elements.

(I hope I dualized it correctly, these results are usually stated for downwards cofinal subsets of $B\bez\{0\}$.)