Is it not enough to modify Peano's construction? An idea (which is different from the onw linked by Iii) might be the following: Peano's construction makes use a function $succ(n)$ which verify the classical properties:

1. There is no $n$ such that $0=succ(n)$
2. $succ(n)=succ(m)$ implies $n=m$
3. If $0\in A$ and $succ(n)\in A$ for all $n\in A$, then $A=\mathbb N$

Maybe it is possible characterize $\mathbb Z$ making use of two functions,, $prec(\cdot)$ and $succ(\cdot)$, related by $prec(succ(n))=succ(prec(n))=n$. Of course, now the first property cannot be true, the second property above has to be required for both $prec$ and $succ$ and, finally, the third property has to be replaced with the following

Induction on $\mathbb Z$: If $A\subseteq Z$ contains at least one element and, moreover, for any $a\in A$ one has $prec(a),succ(a)\in A$, then $A=\mathbb Z$.

Should work.

Is it not enough to modify Peano's construction? An idea (which is different from the onw linked by Iii) might be the following: Peano's construction makes use a function $succ(n)$ which verify the classical properties:

1. There is no $n$ such that $0=succ(n)$
2. $succ(n)=succ(m)$ implies $n=m$
3. If $0\in A$ and $succ(n)\in A$ for all $n\in A$, then $A=\mathbb N$

Maybe it is possible characterize $\mathbb Z$ making use of two (different) functions, $prec(\cdot)$ and $succ(\cdot)$, related by $prec(succ(n))=succ(prec(n))=n$. Of course, now the first property cannot be true, the second property above has to be required for both $prec$ and $succ$ and, finally, the third property has to be replaced with the following

Induction on $\mathbb Z$: If $A\subseteq Z$ contains at least one element and, moreover, for any $a\in A$ one has $prec(a),succ(a)\in A$, then $A=\mathbb Z$.

Should work.