Found this on Weakly Clean Rings and Almost Clean Rings by Ahn & Anderson.
Definition 2.1. A ring $R$ is almost clean if each $x\in R$ can be written as $x=r+e$ where $r\in reg(R)$, the set of regular elements of $R$, and $e\in Id(R)$.
Let $R$ be a commutative ring and $M$ an $R$-module. The idealization of $R$ and $M$ is the ring $R(M)=R⊕M$ with product $(r,m)(r',m')=(rr',rm'+r'm)$.
Theorem 2.11. Let $R$ be a commutative ring and $M$ an $R$-module. Then the idealization $R(M)$ of $R$ and $M$ is almost clean if and only if each $x\in R$ can be written in the form $x=r+e$ where $r\in R−(Z(R)∪Z(M))$ and $e\in Id(R)$.
Proof. We first observe that $Id(R(M))=\{(e,0)\in R(M)|e\in Id(R)\}$.
Suppose $(e,m)\in Id(R(M))$; so $(e,m)=(e,m)^2=(e^2,2em)$. Hence $e=e^2$ and $m=2em$. So $em=2e^2m=2em$ gives $em=0$ and hence $m=2em=0$.
As the other containment is obvious, we have equality.
Also, if $(r,0)\in reg(R(M))$, then $r\in R−(Z(R)∪Z(M))$. For if $r\in Z(R)$, then $rs=0$ where $s=0$ and then $(r,0)(s,0)=(0,0)$; while if $r\in Z(M)$, then $rm=0$ where $m=0$ and then $(r,0)(0,m)=(0, 0)$.
Conversely, if $r\in R−(Z(R)∪Z(M))$, then $(r,m)$ is regular. For $(r,m)(s, n)=(0, 0)$ gives $rs=0$ and hence $s=0$ and then $rn=0$ and hence $n=0$.
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What does "As the other containment is obvious, we have equality." means (or refers to)?
It means “everything of the form $(e,0)$ is obviously in $Id(R(M))$, so that the equality
$Id(R(M))=\{(e,0)\in R(M)|e\in Id(R)\}$
is proven.”