I have seen that to calculate the fundamental group of $\pi_1(S^1)$ one does something like what is presented in the link https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Dooley.pdf ,page 9.
Is there any other way to find the group?
It doesn't matter if it's more laborious, how would it be?
This answer invokes two facts about homology groups and the concept of cogroup and group objects. It is probably not as transparent as the proof based on the covering projection $\mathbb R \to S^1$.
(1) $H_1(S^1) \approx \mathbb Z$.
(2) Let $X$ be a path connected space. Then the abelianization $\pi_1 (X,x_0)_{ab}$ of $\pi_1 (X,x_0)$ is isomorphic to $H_1(X)$. See e.g. The First Homology Group is the Abelianization of the Fundamental Group.
Thus it suffices to show that $\pi_1(S^1,1)$ is abelian.
As Connor Malin mentions in his answer, $(S^1,1)$ is both a group object and a cogroup object in the homotopy category of pointed spaces. It is a well-known result for general categories that if one has a cogroup object $C$ and a group object $G$, then the set of morphisms $Hom(C,G)$ has two group structures, one induced by the comultiplication on $C$ and the other by the multiplication on $G$, which agree and are abelian.
Apply this to $\pi_1(S^1,1) = [(S^1,1),(S^1,1)]$ and note that the group structure of fundamental groups is induced by the comultiplication on $(S^1,1)$.
Here is a way though it is rather convoluted:
I list some standard facts:
The standard way of showing that the fundamental group is a group essentially shows that $S^1$ is a cogroup object in the category of topological spaces. A cogroup object is defined to be a group object in the opposite category.
$S^1$ is also a group object, it is literally a topological group: the units in the complex numbers.
It is easy to prove that the fundamental group preserves products.
Van Kampen's theorem, which I believe does not rely on the fundamental group of the circle, tells us that for reasonably nice spaces the fundamental group also preserves coproducts (recall a coproduct of pointed spaces is the wedge and a coproduct of groups is the free product).
The initial and final object in the category of pointed spaces is a point. The fundamental group of a point is trivial. Since the initial and final object in Grp is a point, the fundamental group preserves initial and final objects.
Now a functor that preserves products and terminal objects sends group objects to group objects. A functor that preserves coproducts and initial objects sends cogroup objects to cogroup objects.
The group objects in Grp are abelian groups. The cogroup objects are the free groups.
Easily follows that the objects that support both a group and a cogroup structure in Grp are $\mathbb{Z}$ and $0$.
So we can say the fundamental group of $S^1$ is either trivial or $\mathbb{Z}$.
If the fundamental group were trivial, then the space would be contractible since the identity resides in the fundamental group. However, the homology of $S^1$ is nontrivial.
We conclude that $\pi_1(S^1)=\mathbb{Z}$.
Since 1965 or so I have preferred the argument using a van Kampen Theorem for the fundamental groupoid $\pi_1(X,S)$ on a set $S$ of base points - see this mathoverflow discussion and the book Topology and Groupoids, which also uses groupoids to give an algebraic model of covering maps by covering morphisms of groupoids, and of actions of a group $G$ on a space by actions on $\pi_1(X,S)$ when $S$ is invariant under the action of $G$ on $X$.
For example if $X$ is the unit interval $[0,1]$ then the groupoid $\mathsf I = \pi_1(X, \{0,1\})$ is a non trivial groupoid which is a generator for the category of groupoids, and if you identify $0,1$ to a single point $\mathsf I$ gets turned into the group of integers. If you do want one base point, which should you choose, and why?
Grothendieck in his 1984 "Esquisse d'un programme" (Section 2) wrote (English translation):
" .., people still obstinately persist, when calculating with fundamental groups, in fixing a single base point, instead of cleverly choosing a whole packet of points which is invariant under the symmetries of the situation, which thus get lost on the way."
Later I ought to add to the above that I feel it is important for students of algebraic topology to be helped to be aware of what is available, and to be encouraged to make their own evaluations. The future belongs to them. I also like the remarks of Gian-Carl Rota advertised here.
