Why is the set of integers with the operation of addition considered a cyclic group?

It is the element $-1$.

On a more serious note, the definition of "generates" includes allowing the inverse of the generating elements. For any group $G$, and element $g\in G$, the subgroup generated by $g$ is $$\{g^n:n\in\mathbb{Z}\}$$ not $$\{g^n:n\in\mathbb{N}\}$$ (the latter is not a subgroup unless $g$ has finite order).

Observe that $g^{-1}$ is always in $\{g^n:n\in\mathbb{Z}\}$.

It is the element $1$ (or the element $-1$).

Why? The subgroup $\langle g\rangle\subset G$ generated by an element $G$ is defined as the smallest subgroup to contain $g$. Since $1$ is in $\langle 1\rangle$ (in $\mathbb Z$) and any subgroup is closed under inverses, $-1$ is also in $\langle 1\rangle$ (since it is the inverse of $1$). Clearly all positive integers are there, and so are their inverses. You get $0$ by the identity group axiom (that the additive identity must be an element of your additive subgroup).