The following are the Peano Postulates:
There exists a set $\mathbb{N}$ with an element $1 \in \mathbb{N}$ and a function $s:\mathbb{N} \to \mathbb{N}$ that satisfy the following three properties.
a. There is no $n \in \mathbb{N}$ such that $s(n)=1$.
b. The function $s$ is injective
c. Let $G \subseteq \mathbb{N}$ be a set. Suppose that $1 \in G$, and that if $g \in G$ then $s(g) \in G$. Then $G=\mathbb{N}$.
We then need to prove this lemma:
Let $a \in \mathbb{N}$. Suppose that $a \neq 1$. Then there is a unique $b \in N$ such that $a=s(b)$.
So, the proof begins by proving that if there were such a $b$, then $b$ is unique.
Then, to show that $b$ exists, we just need to define $G=\{1\} \cup \{c \in \mathbb{N} |$there is some $b \in \mathbb{N}$ such that $s(b)=c \}$ and prove that $G=\mathbb{N}$ because this will immediately imply the existence part of the lemma.
I don't understand how proving that $G=\mathbb{N}$ will imply that $b$ exists. What is the idea that this proof is using to prove that $b$ exists and how is it dependent on $G=\mathbb{N}$?
Actually it is not so good idea to use the same letter $b$ for these different numbers. $b$ (replace it with $d$, for instance) in the definition of $G$ is not the same $b$ from your lemma. $G=\mathbb{N}$ means that each number except $1$ is a successor of some number. Therefore $a$ is also a successor for some number $b$.