Do you know this finitely presented group on two generators?

Note that the abelianized presentation is $langle a,bmid ab^2rangle$. This suggest changing generators so that $ab^2$ is a generator. Define $x=ab^2$, $t=ab$, so that $a=tx^{-1}t$, $b=t^{-1}x$. This yields the presentation $$G=langle t,xmid x(t^{-2}xt^2)(t^{-1}xt)^{-1}rangle.$$ Write $y=t^{-1}xt$. Then this yields: $$G=langle t,x,ymid t^{-1}yt=x^{-1}y,; t^{-1}xt=yrangle.$$ Since $(x^{-1}y,y)$ is a basis of the free group on $x,y$, we identify a semidirect product of the free group $F(x,y)$ on $x,y$ by a cyclic group $langle trangle$ acting on $F(x,y)$ through powers of the automorphism $(x,y)mapsto (x^{-1}y,y)$. (In particular, $G$ is torsion-free.)

Note that this automorphism is not inner, but its square is $(x,y)mapsto ((x^{-1}y)^{-1}y,y)=(y^{-1}xy,y)$, which is an inner automorphism (right conjugation by $y$). Hence the unique subgroup of index 2 in $G$ is a direct product $F(x,y)timeslangle t^2y^{-1}rangle$. Note that the element $t^2y^{-1}$, whose centralizer is this subgroup of index 2, is just equal to $a$.