So the group $(G, \circ)$ must obey at least 3 properties according to my textbook (I am interested only in the third one, so won't give a precise definitions of the first two): 1) binary operation $\circ$ is associative; 2) existence of an identity element such that it does not change any $g \in G$ when composed via $\circ$ operation in either way; 3) $\forall g \in G, \exists g^{-1} : g \circ g^{-1} = id_G = g^{-1} \circ g$.
So, $(\mathbb{Z}, +)$ is a valid group according to my textbook. Obviously, $0 \in \mathbb{Z}$. but there is no distinguishable $0^{-1}$ element such that $0 + 0^{-1} = 0 = 0 + 0^{-1}$. So either the group's definition given by my textbook is incomplete, or $g \in G$ and $g^{-1} \in G$ might be the same element?
Yes, it is allowed for $g$ and $g^{-1}$ to be the same element, and that's exactly what happens here.
In general, when mathematicians say "there exists $x$ with [some property]", it is okay for $x$ to be something that's already been mentioned. If the textbook wanted to say that $g$ and $g^{-1}$ have to be different, it would say, \begin{equation*} \forall g \in G, \exists g^{-1} : g \circ g^{-1} = id_G = g^{-1} \circ g\ \text{and}\ g^{-1} \neq g \end{equation*}