Is every idempotent semiring a completely regular?

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A semiring $(S, +, \cdot)$ is said to be a completely regular semiring if for every $a\in S$, there exists some $x\in S$ satisfying the following conditions: (1) $a=a+x+a $ (2) $a+x=x+a$ and (3) $a(a+x)=a+x$.

In view of the above definition, it looks like every Idempotent semiring is a completely regular because in general a semiring is considered to be a commutative monoid with respect to addition and a semigroup with respect to multiplication, and additive identity is multiplicatively an absorbing element. For instance, if $0$ is the additive identity (i.e., $a+0=0=0+a$) and multiplicatively an absorbing (i.e., $a\cdot 0=0=0\cdot a$) for all $a\in S$. Then the above definition will be satisfied when we take $x=0$. Please someone correct me if i am wrong.

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If an "idempotent semiring" means one in which both operations are idempotent, then both $x=0$ or $x=a$ will work.

Without idempotency of the multiplication operation, it is hard to see how to satisfy $a(a+x)=a+x$. With either $x=0$ or $x=a$, that would amount to $a^2=a$.

Choose any ring with an ideal $A$ such that $A^2=\{0\}$. For example, you could take $R=\mathbb R[x]/(x^2)$ and let $A=(x)/(x^2)$.

The lattice of ideals of $R$ forms an idemoptent semiring, but

$A(A+A)=AA=0$ and $A+A=A\neq\{0\}$.