It is known that, for a boolean nonzero ring (i.e. all elements are idempotent), the subiacent additive group is a group of exponent $2$ (thanks to @Arturo for pointing out, a group of exponent $2$ is a group where every nonzero element has order $2$, that is $x+x=0,\forall x$.). That's because any boolean nonzero ring has characteristic $2$.
How about the converse? I am posing the following question:
Given $(A,+)$ a group of exponent $2$ (which is abelian by default), can we put a multiplication $\cdot:A\times A\rightarrow A$ on it such that $(A,+,\cdot)$ becomes a boolean ring?
I know that, if $A$ is finite, then $|A|=2^t$ for some $t\in\mathbb{N}$ (due to Lagrange and Cauchy's theorems), and moreover there is a group isomorphism $A\cong\mathbb{Z}_2^t$ (use induction on $t$, and take the quotient by any nonzero element). Since $\mathbb{Z}_2^t$ is a product of boolean rings, hence boolean, we can transport the multiplication via isomophism, to make $A$ a boolean ring.
If $A\cong \prod_{i\in I}\mathbb{Z}_2$ for some set $I$, the above reasoning tells us that $A$ is also boolean, with the induced multiplication. This is not generally true, for instance $(\mathbb{N},\text{xor,and})$ is boolean (although not unital), with the operations taken on bits. It doesn't admit such a factorization because the cardinality of $\mathbb{N}$ is not a power of two.
Any further ideas or references are welcome.
As @Tobias Kildetoft suggested, any abelian group $(A,+)$ where any nonzero element has order $p$, where $p$ is a prime, can be endowed with a $\mathbb{Z}_p$-vector space structure, with multiplication naturally defined as follows: $\widehat{a}x:=ax,\forall a\in\overline{0,p-1},x\in A$. We use this fact for $p=2$.
Now, take $B\subseteq A$ a basis, and define $\cdot:A\times A\to A$ as follows: for $x,y\in A$, we have the unique decompositions: $x=b_1+...+b_r,y=c_1+...+c_s$, with all $b_i,c_j\in B$. There is here a good analogy with the idea of bits. Notice that $x+y$ is precisely the sum of base elements they do not have in common, some sort of "xor".
Now set: $$x\cdot y:=\sum_{i=1}^r\sum_{j=1}^sb_i\cdot c_j,$$ where $b_i\cdot c_j:=b_i$ if $b_i=c_j$, and $0$ otherwise. In other words, $x\cdot y$ is the sum of base elements they have in common, exacly what "and" on bits does!
We claim that $(A,+,\cdot)$ is a boolean ring, hence the conclusion. This should be similar to the proof that $(\mathbb{N},\text{xor,and})$ is boolean. Note, however, that $A$ is unital if and only if $\dim A$ is finite, that is, $A$ is finite. In this case, $1_A=\sum_{b\in B} b$.
We can give a shorter proof. In fact, $(A,+,\cdot)$ constructed above is isomorphic to the subring of $\mathbb{Z_2}^B$ consisting of sequences with only finitely many components equal to $1$. As it is a subring of a boolean ring, it is, of course, boolean.
Generalization. As noted before, consider $(A,+)$ an abelian group, with all nonzero elements having order $p$=prime. View $A$ as a $\mathbb{Z}_p$-vector space. Given $B$ a basis, we conclude that $(A,+)$ is isomorphic with the subgroup of $\mathbb{Z}_p^B$ consisting of sequences with only finitely many nonzero components. Transporting the structure via isomorphism, we get that $A$ has a structure of a ring with characteristic $p$, with the property: $x^p=x,\forall x\in A$.
If $B=(b_i)_{i\in I}$, the multiplication on $A$ is : $\sum_{i\in I} x_ib_i\cdot \sum_{i\in I}y_ib_i:=\sum_{i\in I}(x_iy_i)b_i$. In this fashion, it is easy to see that $A$ has a structure of an associative $\mathbb{Z}_p$-algebra, which is unital if and only if $A$ is finite.