I answered my own question. As usual, I was overthinking. In writing this up, I figured it out, so thought I'd keep it as a Q&A, seeing as though I couldn't find it anywhere.
The Question:
Show that every Boolean ring is regular.
The Details:
Definition: A ring $R$ is Boolean if $a^2=a$ for all $a\in R$.
All Boolean rings are commutative.
Definition: An element $x$ in a ring $R$ is regular if there exists a $y\in R$ such that $x=xyx$. We call $R$ regular if all its elements are regular.
Thoughts:
Let $R$ be Boolean, $a,b\in R$. Then
$$aabb=a^2b^2=ab=(ab)^2=abab.$$
What now?
Further Context:
I wrote my undergraduate dissertation on inverse semigroups, which are regular semigroups whose idempotents commute. This ought to be something I can answer myself. I gave it a while but . . .
Let $x\in R$ for a Boolean ring $R$. We have
$$\begin{align} x&=x^2\, \text{(Boolean)}\\ &=xx\, \text{(Indices)}\\ &=xx^2\, \text{(Boolean)}\\ &=x\color{red}{x}x\, \text{(Indices)}. \end{align}$$
So just let $y=\color{red}{x}$.