Do Boolean rings always have a unit element?

An example: the family of all finite subsets of a given infinite set.

In the ring $\prod_{i=1}^\infty\Bbb Z_2$, consider the ideal $\oplus_{i=1}^\infty\Bbb Z_2$. It is a subrng without identity.

(Ultimately I think this is isomorphic to Boris' example, but here the operations are clear.)

Here is a positive result: Let $R$ be a commutative ring such that every element can be written as a sum of products of elements of $R$ (i.e. the multiplication map $R \otimes_{\mathbb{Z}}R \to R$ is surjective). If $R$ is finite, then $R$ is unital. In particular, every finite boolean ring is unital (which also can be proven directly, of course).

Proof: Consider the unitalization $R^+$ as an $R$-module. Then we have $R R = R$, hence by Nakayama there is some $e \in R$ with $(1-e) R=0$. But this means that $e$ is a unit of $R$.