Let $G$ be a group and denote by $G_2$ the set of all $x$ such that $x^2=1$. By a positive-negative partition of $G$, I mean a disjoint union $G=G_+\dot{\cup} G_- \dot{\cup} G_2$ such that $G_-=(G_+)^{-1}:=\{x^{-1}: x\in G_+\}$.
Positive-negative partitions of groups always exist. Define the equivalent relation $\sim$ by $x\sim y$ if and only if $x\in\{y,y^{-1}\}$. Then $G\setminus G_2$ is either empty or has a partition with two members classes induced by $\sim$. Now, define $G_+$ as an arbitrary transversal of the partition of $G\setminus G_2$ if $G\neq G_2$, otherwise $G_+=\emptyset$, and put $G_-:=(G\setminus G_2)\setminus G_+$. Then $G=G_+\dot{\cup} G_- \dot{\cup} G_2$ is a positive-negative partition for $G$. It is easy to see that $G_+$ is a sub-semigroup of $G$ if and only if $G_-$ is so.
For example, the set of natural numbers $\mathbb{N}$ (positive integers) is only one choice among an infinite number of such subsets $\mathbb{Z}_+$. Also, $G=\mathbb{Z}$ has the unique representation $G=G_+\dot{\cup} G_- \dot{\cup} G_2$ (up to the order of the subsets) under the condition that $G_+$ is a sub-semigroup.
َAlso, if $G$ is finite, then $G_+$ or $G_-$ can not be a sub-semigroup of $G$ (since every sub-semigroup of a finite group is a subgroup).
Now, let $G$ be a group containing no nontrivial elements of finite order, then:
(1) Is it true that there is a $G_+$ being a sub-semigroup of $G$?
(2) If yes, is it unique (under some conditions)?
(3) Do you know of a related topic and/or reference?
Thanks in advance.
You have described one of the equivalent definitions of a left-orderable group. See https://en.wikipedia.org/wiki/Linearly_ordered_group. Some examples of torsion-free non-left-orderable groups are listed. If a group is left-orderable, the ordering is generally very far from unique, e.g., in $\mathbb Z^2$ you can consider any irrational half-plane.