Here is a statement I am trying to prove, that will allow me to prove the inverse exists in the group of units of the additive group of modulo n.
Let $n\in \mathbb{N}$ and $a\in\mathbb{Z}$. If gcd$(a,n) = 1$, there exists $0\leq b < n$ such that $ab \equiv 1$ (mod $n$), and gcd$(b,n) = 1$.
The proof I can think of is this:
Since $(a,n) = 1$, there exists some $c_1$ and $c_2 \in \mathbb{Z}$ such that $c_1a+c_2n = 1$. Then $c_1a = -c_2n + 1$, and this means $c_1a \equiv 1$ (mod $n$). Consider the ideal $J$ generated by $c_1$ and $n$, which is J := $\{x_1c_1 + x_2n : x_1, x_2 \in \mathbb{Z} \}$, then $1 \in J$, and since $1$ is the smallest positive integer, it must be the generator for $J$ and hence it is the greatest common divisor for $c_1$ and $n$.
But this proof does not consider the inequality. So my question is: from where does the inequality arise?
I don't understand the problem. You already proved that there exists $b\in\mathbb{Z}$ such that
$ab\equiv 1$(mod $n$) and $gcd(b,n)=1$. If your question is why $0\leq b<n$ then it just follows from the fact that you can always take $0\leq c<n$ such that $c\equiv b$(mod $n$) and then $ac\equiv ab$(mod $n$). So without loss of generality you can say $b$ itself is in $\{0,1,...,n-1\}$.