See the answer of Ian Agol to a question of mine. mathoverflow.net/questions/200080/…

The problem is open. See e.g. Carter, William. New examples of torsion-free non-unique product groups. J. Group Theory 17 (2014), no. 3, 445--464.

See above for an equivalent question

@MielSharf: Could you please let us know on which page of Woess' article your claim is mentioned (or is used)? For upper central series is true. As every f.g. nilpotent group has a torsion-free nilpotent subgroup of finite index.

@YCor: Thanks. It is also easy to prove the non-existence of such torsion-free groups if they satisfy unique product property. In the meanwhile, if such group exists the size of $B$ must be even.

My last comment: So we have in hand a torsion-free non-left orderable group defining by two generators and two relations. So the group may have Unique Product property?! If not, we have also a surprising group.

I do not say it must be "finitely presented" only I am surprised. Don't mind! Sorry.

@IanAgol: For any torsion-free non-left-orderable factor $B_3/N$ of $B_3$, we know that $N$ must be a free subgroup of $B_3$ such that $N\cap Z(B_3)=1$. It is a necessary condition. So it does not seem (or at least surprising) one can define a torsion-free non-left-orderable factor of $B_3$ with finitely many relations.

@IanAgol: Sorry to ask further. You mean that to define $\pi_1(S_r^3(T))$ one needs only 2 relations (one of Braid group $B_3$ i.e. $xyx=yxy$ and another one, that you explain above) or the relation you gave is a series of relations. I can ask it as a new question.

@IanAgol: I mean for those $r$ such that $\pi_1(S_r^3(T))$ is not right orderable and is torsion-free.

@IanAgol: Is it known a group presentation (not necessarily finite) for $\pi_1(S_r^3(T))$ for some $r$ with $|r|\geq 1$?

@YCor: Could you please kindly give a reference for the being open of the case torsion-free finite-index subgroups of $SL_3(\mathbb{Z})$?