If $R$ is a commutative unital ring with no non-trivial idempotents, then is the group of units of $R$ not isomorphic with the additive group of $R$? 
Let $R$ be a commutative unital ring such that $R$ has no non-trivial idempotents, and let $(R^\times,\ \cdot)$ denote the group of units of $R$. Is it true that $(R,+) \ncong (R^\times,\ \cdot)$ ? 

See Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?. The counter-examples (as far as I can see) do not cover this. Please help. Thanks in advance.
 A: Before I present my counterexample, let me point out that this question shows that $R$ cannot be taken to be a field. As you pointed out in the question, the solution is also known in the case when non-trivial idempotents are allowed. My answer to your question is a bit more involved.
My construction depends on a field $K$ and a group $G$. I denote $$\mathcal{H}(K,G)=\left\{\varphi+f\left|\right.\varphi\in KG,\,\,f\in t\,K[t]\right\},$$ that is, $\mathcal{H}(K,G)$ is the set of all formal sums $\varphi+f$ in which $\varphi$ is an element of the group ring $KG$ and $f$ is a polynomial with zero constant term. I define $$(\varphi_1+f_1)(\varphi_2+f_2)=\varphi_1\varphi_2+\sigma(\varphi_1) f_2+\sigma(\varphi_2) f_1+ f_1 f_2,$$ where $\sigma:KG\to K$ is the mapping sending an element of a group ring into the sum of its coefficients.
Let me omit a formal proof that $\mathcal{H}(K,G)$ is indeed a ring. It is also clear that the invertible and idempotent elements in $\mathcal{H}(K,G)$ are precisely those that live in $G$.
To construct the counterexample, I set $K=\mathbb{F}_2$ and $G=\mathbb{Z}_2^\infty$. We get $\varphi^2=\sigma(\varphi)$ for all $\varphi\in KG$, so that there are no non-trivial idempotents, and every invertible element squares to $1$. Therefore, the multiplicative subgroup of $\mathcal{H}(K,G)$ is isomorphic to the direct sum of several copies of $\mathbb{Z}_2$. The same holds for the additive group of $\mathcal{H}(K,G)$ because the underlying field has characteristic two. It remains to note that the cardinalities of the multiplicative and additive groups of $\mathcal{H}(K,G)$ are equal.
