Let $\{R_α\}$ be a nonempty collection of commutative rings.
Then each $x\in R=∏R_α $ can be written as $x=r+e$ where $r\in reg(R)$ (the set of regular elements of
R) and $e\in Id(R)$ (the set of idempotent elements of R) if and only if for each $R_α$, every $x_α\in R_α$ can be written as $x_α=r_α+e_α$ where $r_α\in reg(R_α)$ and $e_α\in Id(R_α)$.
This is Proposition 2.10 in "Weakly Clean Rings and Almost Clean Rings" by Ahn & Anderson (2006).
How to prove that?
By homomorphism we can get the idempotent in $R=∏R_α $ and $R_α$. I don't have any idea how to work with the regular elements.
Here is how homomorphism works for idempotent element:
There is a homomorphism mapping $π_α:R→R_α$ that is $π_α(x)=x_α$, with $x=(x_a)∈∏R_α=R$.
We know that homomorphism preserve operation :
For all $x,y∈R$,
$π_α(x+y)=π_α (x)+π_α (y)$ and $π_α(xy)=π_α (x) π_α (y)$.
Then we know that the image of an idempotent element in $R=∏R_α$ id is an idempotent in $R_α$.
For all $e=(e_α)∈Id(R)$ we can get $(π_α (e))^2=π_α (e^2 )=π_α (e)$
These homomorphisms $\pi_\alpha$ are called projections. With your homomorphism-argument you proved that if $e$ is an idempotent element of $R$ then $\pi_\alpha(e)$ is a idempotent element in $R_\alpha$. But you did not prove the opposite direction: if $e_\alpha$ is and idempotent element of $R_\alpha$, for all $\alpha$, then $(e_\alpha)_\alpha$ is and idempotent element of $R$.
Instead of investigating the homomorphisms further let's use the definition of the direct product.
Remember that notations like $\{R_\alpha\}$ and $\Pi R_\alpha$ are shortcuts and hide that there must be a set where the $\alpha$ is from, the index set. So lets call this set $A$. So $\{R_\alpha\}$ actually is $\{R_\alpha\}_{\alpha \in A}$ or $\{R_\alpha\mid \alpha \in A\}$ and $\Pi R_\alpha$ is $\Pi_{\alpha \in A} R_\alpha$. An element $r \in \Pi_{\alpha \in A} R_\alpha$ is a function $r:A\mapsto R$ and for this function we use the notation $(r_\alpha)_{\alpha\in R}$ or $(r_\alpha)_\alpha$ for short, too. The sum $r^{(1)}+r^{(2)}$ is the functions that maps $\alpha$ to $r^{(1)}(\alpha)+r^{(1)}(\alpha)$, $^{(1)}\cdot ^{(2)}$ is defined ins a similar way.
If $e\in R$ we write $(e_\alpha)_\alpha$ for it. If it is an idempotent element we have $$(e_\alpha)_\alpha=(e_\alpha)_\alpha\cdot(e_\alpha)_\alpha \tag 1$$ Fom the definition of $\cdot$ in $R$ follows $$(e_\alpha)_\alpha\cdot(e_\alpha)_\alpha = (e_\alpha\cdot e_\alpha)_\alpha\tag 2$$ an so $$(e_\alpha)_\alpha=(e_\alpha \cdot e_\alpha)_\alpha\tag 3$$ which is equivalent to $$\forall \alpha\in A\!:\; e_\alpha=e_\alpha\cdot e_\alpha\tag 4$$ This is what you have already proved.
Now we prove the opposite direction. It is easy to see that from $(4)$ follows $(3)$ and that $(2)$ holds. From $(2)$ and $(3)$ follows $(1)$.
So we have proved that $(e_\alpha)$ is idempotent if and only if for all $\alpha$ is $ e_\alpha$ idempotent.
The $0$-element of $R$ is $(0)_\alpha$. If $(r_\alpha)_\alpha\in R$ is a regular element then for each $\alpha$ the element $r_\alpha$ is a regular element in $R_\alpha$. Otherwise if $r_{\alpha_0}$ is a zero divisor, than $r_{\alpha_0}\cdot b=0$ for a $b \in R_{\alpha_0}$ and $b\ne0$. If we set $$c_\alpha=b, \alpha=\alpha_0$$ $$c_\alpha=0, \alpha\ne\alpha_0$$ then $(c_\alpha)_\alpha\ne 0$ and $(r_\alpha)_\alpha\cdot (c_\alpha)_\alpha=(0)_\alpha$, so $r$ would not be regular. If $r_\alpha\in R_\alpha$ is regular for each $\alpha$ then $(r_\alpha)_\alpha$ is regular, too.
Now similar statement for a decomposition of a ring element $x$ into $e+r$, where $e$ is idempotent and $r$ is regular follows immediately.
Note: For zero divisor the following holds: If $n_\alpha)_\alpha$ is a zero divisor of $R$ if and only if there exists at least one $\alpha$ such that $n_\alpha$ is a zero divisor in $R_\alpha$.
In symbolic notation
$$(r_\alpha)_\alpha \in \text{Idempotent}(\Pi_\alpha R_\alpha)\Longleftrightarrow \forall \alpha\!:r_\alpha \in \text{Idempotent}(R_\alpha)$$ $$(r_\alpha)_\alpha \in \text{Regular}(\Pi_\alpha R_\alpha)\Longleftrightarrow \forall \alpha\!:r_\alpha \in \text{Regular}(R_\alpha)$$ $$(r_\alpha)_\alpha \in \text{Zerodivisor}(\Pi_\alpha R_\alpha)\Longleftrightarrow \exists \alpha\!:r_\alpha \in \text{Zerodivisor}(R_\alpha)$$ The following holds per definition of "regular" $$(r_\alpha)_\alpha \in \text{Regular}(\Pi_\alpha R_\alpha)\Longleftrightarrow \lnot((r_\alpha)_\alpha \in \text{Zerodivisor}(\Pi_\alpha R_\alpha))\tag{11}$$ This is the property of Zerodivisors we have to prove: $$\lnot((r_\alpha)_\alpha \in \text{Zerodivisor}(\Pi_\alpha R_\alpha))\Longleftrightarrow \lnot(\exists \alpha\!:r_\alpha \in \text{Zerodivisor}(R_\alpha))\tag{12}$$ and the right hand site term is logically equivalent to $$ \lnot(\exists \alpha\!:r_\alpha \in \text{Zerodivisor}(R_\alpha))\Longleftrightarrow \forall \alpha\!:\lnot (r_\alpha \in \text{Zerodivisor}(R_\alpha))\tag{13}$$ and again by the definition of "regular" we get $$\forall \alpha\!:\lnot (r_\alpha \in \text{Zerodivisor}(R_\alpha))\Longleftrightarrow\forall \alpha\!:r_\alpha \in \text{Regular}(R_\alpha)\tag{14}$$ To prove $(12)$ you prove $$(r_\alpha)_\alpha \in \text{Zerodivisor}(\Pi_\alpha R_\alpha)\Longleftrightarrow \exists \alpha\!:r_\alpha \in \text{Zerodivisor}(R_\alpha)\tag{15}$$ I hope it is clear how to do this