Any $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has  a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.


 Moreover, if a  0-simple semigroup $S$ with 1 has no other idempotents, then it is a group with 0. 

*Proof:* Let $G$ be its subgroup of invertible elements. For every $a\in S\setminus 0$ there are such $x,y\in S$ that $xay=1$. Then $(ayx)^2=ayx$ whence $ayx=1$. Since $xay=ayx=1$, hence $x\in G$. But then $a=x^{-1}y^{-1}\in G$, i.e. $S=G\cup 0$.