Van Kampen 's Theorem :
Let $X= A_1 \cup A_2$,where $A_1 ,A_2 $ and $A_1 \cap A_2 $ are path connected and let $x_0 \in A_1\cap A_2 $ then $\pi_1 (X,x_0) = \frac{\pi_1(A_1) * \pi_1(A_2) }{N}$
where $N$ is the smallest normal subgroup generated by the elements $[(j_1i_1)_*(w)]*[(j_2i_2)_*(w)]^{-1}$ where $w\in \pi_1(A_1 \cap A_2)$
Compute $\pi_1(S^2)$
My attempt :- $A_1=S^2\setminus \{n\} $ and $A_2=S^2\setminus\{s\} $ where $n$ and $s$ denote the north and south poles respectively
$\pi_1(A_1)=\pi_1(A_2)= \{e\}$
$\pi_1(A_1 \cap A_2)= \mathbb{Z}$
By Van Kampen's thorem $\pi_1 (S^2,x_0) = \frac{e * e }{N}$
My question : How to find the value of $N$ ?
My attempt :we know that $N$ is the smallest normal subgroup generated by the elements $[(j_1i_1)_*(w)]*[(j_2i_2)_*(w)]^{-1}$ where $w\in \pi_1(A_1 \cap A_2)$
Here $(i_1)_*:\pi_1(A_1\cap A_2) \to \pi_1(A_1)$
$(i_2)_*:\pi_1(A_1\cap A_2) \to \pi_1(A_2)$
$(j_1)_*:\pi_1(A_1) \to \pi_1(A_1 \cup A_2)$
$(j_2)_*:\pi_1(A_2) \to \pi_1(A_1 \cup A_2)$
$\implies [(j_1i_1)_*(w)]*[(j_2i_2)_*(w)]^{-1}= [j_1(\pi_1(A_1))(\mathbb{Z})]*[j_2(\pi_1(A_2))(\mathbb{Z})]^{-1}$
After that im not able to proceed further
Notice that $e * e$ is the trivial group, so the only possibility for $N$ is the trivial group. This is the fastest way to identify $N$ from van Kampen's theorem and your set-up.
For some more insight, look at what the maps $i_1$ and $i_2$ do: they map from $\pi_1(A_1 \cap A_2)$ to $\pi_1(A_1)$ and $\pi_1(A_2)$ respectively. You correctly identified the isomorphism types of all these groups, so you can tell $i_1$ and $i_2$ are homomorphisms from an infinite cyclic group to the trivial group. Therefore must both be the trivial homomorphism. What this means geometrically is that any path in $A_1 \cap A_2$ (which has the homotopy type of the circle) becomes nullhomotopic when viewed as a path in either $A_1$ or $A_2$, which are simply connected.