Let $BA$ be the set of axioms of Boolean algebras.
Let $Th(BA)$ be the set of first-order sentences that are true when conjoined with the axioms in $BA$.
I believe that $Th(BA) = Th(\langle \{0,1\}, \land, \lor, \lnot \rangle)$ where $\land, \lor,\lnot$ are the usual propositional operations interpreted over $\{0,1\}$.
Is this result true?
This is false. For example, the sentence $$\forall x,y,z(x=y\vee x=z\vee y=z)$$ is true in the two-element Boolean algebra but false in others.
A less-trivial example: any finite Boolean algebra satisfies the sentence "There is an atom" ($\exists x,y\forall z[x<y\wedge \neg (x<z\wedge z<y)]$) but there are infinite atomless Boolean algebras.
That said, if we restrict attention to equational properties - equational logic being the small fragment of first-order logic consisting of the universal closures of atomic formulas - then we get an affirmative answer, since every Boolean algebra is a quotient of a subalgebra of a power of the $2$-element B.A. and each of those operations on structures preserves equational properties. (The converse of this is deeper but still true - see "HSP theorem.")