Consider the following property for a group $(\mathcal{G},\cdot,1)$: 

> There are exactly three conjugacy classes $\{1\}$, $\mathcal{C}_1$, $\mathcal{C}_2$ in $\mathcal{G}$, and we have $\mathcal{C}_1 \mathcal{C}_1 \subseteq \mathcal{C}_1$ and $\mathcal{C}_2=\mathcal{C}^{-1}_1$.

Note that the only finite groups with exactly three conjugacy classes are the cyclic group of order $3$ and the symmetric group of order $6$. Those do not satisfy the property above. So any such group must be infinite, hence also non-abelian.

In fact, given such a group $\mathcal{G}$, define $x<y$ if and only if $y \cdot x^{-1} \in \mathcal{C}_1$. This defines a linear order on $\mathcal{G}$ for which it is bi-ordered, i.e. with $\forall x,y,z \in \mathcal{G},\ y>1 \Longleftrightarrow x \cdot y  \cdot z> x \cdot z$. So it is perhaps best to think of those groups as linearly bi-ordered groups where any two strictly positive elements are conjugated.

I suspect not so many such groups are known. So my question is: **are there known examples of such groups**? Or better yet: **have they been studied to some extent**?

One could also expect such groups to be related to the first-order theory $T$ of linearly bi-ordered non-trivial groups with trivial center. The question would be: **does any algebraically closed model of $T$ satisfy the property above**?
Indeed I believe that László Fuchs proved that in the case of partially bi-ordered, lattice ordered groups, the algebraically closed models have the property that any two strictly positive elements are conjugated. In that case however there may still be more conjugacy classes than three because not every element must be positive or negative.