In the answer I will use the standard way to number the relators in a presentation. Then, if $G$ is a group of deficiency $\ge 1$ (i.e., admits a presentation with $n$ generators and $k$ relators, where $n>k$) then $G$ is infinite and, moreover, admits an epimorphism to the infinite cyclic group. To prove this, consider the rational vector space $V=Hom(G, {\mathbb Q})$ (where we regard ${\mathbb Q}$ as the additive group of the field ${\mathbb Q}$):
This vector space $V$ is given by imposing $k$ linear equations on
$Hom(F_n, {\mathbb Q})={\mathbb Q}^n$, since every homomorphism to ${\mathbb Q}$ is determined by its values on generators of $G$, while the only restrictions on the images of generators are that each relator maps to zero: Every such condition is one linear equation.
Hence, $dim Hom(G, {\mathbb Q}) \ge n-k\ge 1$. It therefore, follows that there exists a nonzero homomorphism $h: G \to {\mathbb Q}$. The image of this homomorphism is an infinite torsion free finitely generated subgroup (as ${\mathbb Q}$ contains no nonzero finite subgroups), i.e. ${\mathbb Z}^r$ for some $r\ge 1$. Since it is a subgroup of ${\mathbb Q}$, $r=1$. Thus, $G$ admits an epimorphism $h: G\to {\mathbb Z}$, and, hence, is an infinite group. In particular, $G$ contains an element of infinite order (any element $g\in G$ such that $h(g)\ne 0$).
In fact, one can prove more, namely that abelianization of $G$ has rank $\ge n-k$, but we do not need this.
A much more interesting result is due to Baumslag and Pride: They proved that every group of deficiency $\ge 2$ has a finite index subgroup which admits an epimorphism to a nonabelan free group. Such a group is called large. See also http://arxiv.org/pdf/1007.1489.pdf and references there.
If the group $H = \langle S \cup \{t\} \mid R \cup \{tat^{-1}=\theta(a) : a \in A\})$ was finitely presented, then it would be presented using a finite subset of the relation set in the infinite presentation. That is, we would have $H = \langle S \cup \{t\} \mid R \cup \{tat^{-1}=\theta(a) : a \in X\})$ for some finite subset $X$ of $A$.
But this presents the group $G_{*B}$, where $B = \langle X \rangle$. So if $A$ is not finitely generated, then $B$ is a proper subgroup of $A$.
But then, by Britton's Lemma on HNN extensions, if we choose $a \in A \setminus B$, then the element $tat^{-1}\theta(a)^{-1}$ of $G_{*B}$ contains no pinch, and so is not equal to the identity. But it is equal to the identity in $G_{*A}$, contradiction.
Best Answer
Note that the abelianized presentation is $\langle a,b\mid ab^2\rangle$. 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 $$G=\langle t,x\mid x(t^{-2}xt^2)(t^{-1}xt)^{-1}\rangle.$$ Write $y=t^{-1}xt$. Then this yields: $$G=\langle t,x,y\mid t^{-1}yt=x^{-1}y,\; t^{-1}xt=y\rangle.$$ 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 $\langle t\rangle$ 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 $F(x,y)\times\langle t^2y^{-1}\rangle$. Note that the element $t^2y^{-1}$, whose centralizer is this subgroup of index 2, is just equal to $a$.