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
The space of marked groups can be used to show some properties of surface groups. For example, see the following question: Is the center of the fundamental group of the double torus trivial?
I found also some interesting results in the appendix of Model Theory by Hodges.
Consider the following problem: If $A,B,C$ are three groups such that $A \times C \simeq B \times C$; when $A$ and $B$ are isomorphic? For example:
See Cancellation and Elementary Equivalence of Finitely Generated Finite-By-Nilpotent Groups or Cancellation of abelian groups of finite rank modulo elementary equivalence by Francis Oger.
Shelah solved the following problems using model theory:
See Uncountable groups have many nonconjugate subgroups.
See On a problem of Kurosh, Johnson groups, and applications.
In On some conjectures connected with complete sentences, Makowski relates the problem "Is there an infinite finitely presented group with a finite number of conjugacy classes?" to the existence of a specific kind of theory (theorem 2.6).
A group $G$ is said linear of degree $n$ if it is embedable into $GL(n,F)$ for some field $F$. In Barwise's book, Handbook of mathematical logic, there is a proof of