The claim that Iwaniec makes is simply false; the mistake originates in Lemma 8 in:
C. L. Siegel, Discontinuous groups. Ann. of Math. (2) 44 (1943), 674--689.
The mistake in the proof is that Siegel incorrectly assumes that every Fuchsian group has a finitely-sided fundamental polygon. (I looked closely in the paper, Siegel's definition of a Fuchsian group only requires discreteness, no assumptions on fundamental polygons.)
The correct result is:
Theorem. The following are equivalent for a Fuchsian group $\Gamma< PSL(2,R)$ of the first kind:
$\Gamma$ is finitely generated.
$\Gamma$ is a lattice in $PSL(2,R)$, i.e. ${\mathbb H}^2/\Gamma$ has finite area.
One (equivalently every) fundamental polygon of $\Gamma$ has only finitely many sides.
See for instance Theorem 10.1.2 in
A.F.Beardon, "Geometry of Discrete Groups", Springer Verlag, 1983.
The example that Lee Mosher gave you is definitely valid although a proof that the uniformizing group $\Gamma$ is of the 1st kind would take some nontrivial effort involving extremal length considerations. An easier example is the following.
Take a torsion-free Fuchsian subgroup $\Gamma_1< PSL(2,R)$ such that ${\mathbb H}^2/\Gamma$ has finite area. Take a nontrivial normal subgroup $\Gamma_2< \Gamma_1$ of infinite index. It is a general fact about general Fuchsian groups that the limit set of a nontrivial normal subgroup $\Gamma_2\triangleleft \Gamma_1$ equals the limit set of $\Gamma_1$. Since our group $\Gamma_1$ is of the first kind, so is the normal subgroup $\Gamma_2$. On the other hand, area is multiplicative under isometric coverings between Riemannian surfaces: If $S_2\to S_1$ is an isometric covering of degree $d$ of Riemannian surfaces, then
$$
Area(S_2)=d Area(S_1).
$$
(The same holds for manifolds in all dimensions, but the area would mean volume.) In our case, the degree of the covering map
$$
{\mathbb H}^2/\Gamma_2\to {\mathbb H}^2/\Gamma_1
$$
equals the index $|\Gamma_1: \Gamma_2|=\infty$. Hence,
$$
Area({\mathbb H}^2/\Gamma_2)=\infty.
$$
Here's a proof. I'm going to represent the hyperbolic plane $\mathbb H$ using the Poincaré disc model $\mathbb D$ instead of the upper half plane model. So with $\mathbb H = \mathbb D$ we also have $\partial\mathbb H = S^1$. Convergence in $\mathbb H \cup \partial \mathbb H = \mathbb D \cup S^1 = \overline{\mathbb D}$ is therefore governed by the Euclidean metric $d_{\mathbb E}$, so it is easier to talk about convergence to boundary points using $\mathbb D$ than it is using the upper half plane model.
Now I'll state a geometric fact that one can use to resolve this problem.
Lemma: For any $z \in \mathbb H$ there exists $r > 0$ such that for any $\gamma \in \Gamma$ there exists $\delta \in \Gamma'$ such that $d_{\mathbb H}(\gamma \cdot z,\delta \cdot z) < r$.
In this statement, the distance function $d_{\mathbb H}$ refers to distance in $\mathbb H$. In words, this lemma says that the entire $\Gamma$-orbit of $z$ is contained in the $r$ neighborhood of the $\Gamma'$ orbit of $z$. This lemma is true for any group acting by isometries on any metric space; I'll leave the proof as an exercise for you to ponder (hint: first pick right coset representatives).
To see how this proves what you want, consider a point $x \in \partial\mathbb H = S^1$. Using that $\Gamma$ is of the first type, choose $z \in \mathbb H$ and a sequence $\gamma_i \in \Gamma$ such that $\lim_{i \to \infty} \gamma_i \cdot x = z$. Applying the lemma, choose $r > 0$ and a sequence $\delta_i \in \Gamma'$ such that $d(\gamma_i \cdot z, \delta_i \cdot z) < r$.
We want to prove that $\lim_{i \to \infty} \delta_i \cdot z = x$. To do this, we use a fact about the Poincaré disc model $\mathbb D$: as the center of a hyperbolic ball of radius $r$ approaches $S^1$, the Euclidean diameter of that ball approaches zero. So, knowing that $\lim_{i \to \infty} \gamma_i \cdot z = x \in S^1$, and that $d_{\mathbb H}(\gamma_i \cdot z,\delta_i \cdot z) < r$, it follows that $\lim_{i \to \infty} d_{\mathbb E}(\gamma_i \cdot z,\delta_i \cdot z ) = 0$. Therefore,
$$\lim_{i \to \infty} \delta_i \cdot z = \lim_{i \to \infty} \gamma_i \cdot z = x
$$
Best Answer
An element of a Fuchsian group is either elliptic, parabolic, or hyperbolic.
If $g$ is elliptic, then the group generated by $g$ is finite.
If $g$ is parabolic then there is only one limit point for $g^n z$, independent of $z$. For example if $g(z) = z+1$, then this limit point is $\infty$ regardless of what $z$ we start at. So then the group generated by $g$ is not Fuchsian of the first kind.
If $g$ is hyperbolic, then there are only 2 limit points for $g^n z$, independent of $z$. For example, if $g(z) = 2z$ then these limit points are 0 and $\infty$. So again the group generated by $g$ is not Fuchsian of the first kind.