Noah's answer is correct: an embedding of atomic Boolean algebras which preserves the atoms is an elementary embedding. Let me give a few more details.
The theory of atomic Boolean algebras, when extended by unary predicates $P_m(x)$ expressing "$x$ has at least $m$ atoms below it" has quantifier elimination (as discussed in your recent question).
Let $h\colon A\to B$ be an atom-preserving embedding of atomic Boolean algebras. I claim that for $a\in A$, $a$ has at least $m$ atoms below it if and only if $h(a)$ does. If $a_1,\dots,a_m$ are distinct atoms below $a$, then $h(a_1),\dots,h(a_m)$ are distinct atoms below $h(a)$ (since $h$ is injective, preserves atoms, and preserves $\leq$). Conversely, if $a$ has fewer than $m$ atoms below it, say $a_1,\dots,a_k$ with $k<m$, then since $A$ is atomic, $a = a_1\vee \dots \vee a_k$, and $h(a) = h(a_1)\vee \dots \vee h(a_k)$ (since $h$ preserves $\vee$). Now if $c\leq h(a)$ is an atom, we have \begin{align*}
c &= c\wedge h(a)\\
&= c\wedge (h(a_1)\vee \dots \vee h(a_k))\\
&= (c\wedge h(a_1))\vee \dots \vee (c\wedge h(a_k)).
\end{align*}
Each $h(a_i)$ is an atom, so $c\wedge h(a_i) = \bot\text{ or }h(a_i)$. And since the $h(a_i)$ are distinct and $c$ is an atom, exactly one of the terms $c\wedge h(a_i)$ is not $\bot$. Thus we have $c = h(a_i)$ for some $1\leq i \leq k$, which shows that there are exactly $k$ atoms below $h(a)$.
What we have shown is that if we expand $A$ and $B$ by the predicates $P_m(x)$, then $h$ is an embedding in the expanded language. By QE, $h$ is an elementary embedding in the expanded language, so it is an elementary embedding in the language of Boolean algebras.
Another perspective on what we've shown here is that the theory of atomic Boolean algebras is model complete in the language where we include a predicate $\mathrm{At}(x)$ naming the atoms. Recall that a theory is model complete if every embedding between models is an elementary embedding. And it's not hard to see that an embedding between atomic Boolean algebras maps atoms to atoms if and only if it is an embedding in the language expanded by $\mathrm{At}(x)$ (this is not immediate: we also need to check that if $a$ is not an atom, then $h(a)$ is not an atom).
A well-known characterization of model complete theories is that $T$ is model complete if and only if every formula is equivalent to both an existential and a universal formula. Now if $T$ has an expansion to a theory with quantifier elimination, and every new atomic formula in the expansion is equivalent to both an existential and a universal formula in the original language, then $T$ is model complete (since any Boolean combination of formulas, each of which is both existential and universal, is both existential and universal).
So we could rephrase the argument above as follows: Show that the theory of atomic Boolean algebras with $\mathrm{At}(x)$ is model complete by showing that all the predicates $P_n(x)$ are both existential and universal in this language.
And indeed, we have that $P_n(x)$ is equivalent to both $$\exists z_1\dots z_n\,\left(\bigwedge_{i\neq j}z_i\neq z_j\land \bigwedge_{i=1}^n (\mathrm{At}(z_i)\land z_i\leq x)\right)$$
and $$x\neq \bot \land \forall z_1\dots z_{n-1}\lnot \left( (z_1\vee \dots \vee z_{n-1} = x)\land \bigwedge_{i=1}^{n-1} \mathrm{At}(z_i)\right).$$
Every Boolean algebra is an algebra of definable sets. More strongly, given any set $X$ and any subalgebra $B\subseteq\mathcal{P}(X)$, there is a first-order structure on $X$ whose algebra of definable subsets (without parameters) is $B$. To do this, just make a unary relation for each element of $B$ which is true on that subset of $X$. Clearly then each element of $B$ is definable. For the converse, any formula can involve only finitely many of the relations, and so it suffices to consider the case where $B$ is finitely generated and hence finite. In that case, $B$ just consists of the unions of elements of some finite partition of $X$. Any two elements of the same block of the partition are indistinguishable with respect to all the unary relations, and so any definable subset must be a union of blocks, and thus an element of $B$.
Best Answer
The trick you're missing is to find an embedding into an atomic Boolean algebra, not a field of sets. Atomic Boolean algebras definitely are the models of a first-order theory (i.e., your assumption holds if yoou replace "field of sets" with "atomic Boolean algebra"), and so you can embed any Boolean algebra into an atomic Boolean algebra. But since any atomic Boolean algebra is a field of sets, this also proves you can embed any Boolean algebra into a field of sets.