Do these axioms fully describe the integers? Here, I use Peano-like axioms to describe the set of integers $Z$. They are based on two successor functions, each starting with a common point of $0$, and a principle of induction for the integers.
Let $Z$, $Pos$, $Neg$, $s$, $s'$ and $0$ be such that:
$Pos\subset Z$
$Neg\subset Z$
$Z=Pos\cup Neg$ (edit)
$\forall x (x\in Pos \wedge x\in Neg \leftrightarrow x=0)$
$s:Pos\rightarrow Pos$
$s$ is injective
$s':Neg\rightarrow Neg$
$s'$ is injective
$\forall x\in Pos (s(x)\neq 0)$
$\forall x\in Neg (s'(x)\neq 0)$
$\forall m ((0\in m\wedge \forall x\in Pos (x\in m\rightarrow s(x)\in m) \wedge \forall x\in Neg (x\in m\rightarrow s'(x)\in m))\rightarrow \forall x\in Z (x\in m)) $
Note that, contrary to the usual convention, I have had to include $0$ in both sets $Pos$ and $Neg$. 
Lemma: $0\in Z, Pos, Neg$
See my follow-up below
 A: Yes, this characterizes the integers as long as the quantifiers on subsets range over all subsets. If you just take the axioms you have for Pos, these give Peano's axioms, which uniquely capture the natural numbers up to isomorphism in full second-order semantics. The same it true for Neg. Thus the overall structure for these axioms will be the integers, up to isomorphism.
Contrary to some claims, it is not very hard to define addition. First, there is a canonical semigroup isomorphism between (Pos, $s$) and (Neg, $s'$) preserving $0$. So this gives a notion of $-x$ for each $x$. Now we only have to define addition for positive numbers, which is described on the Wikipedia article, and then we use the negation operation to define addition for arbitrary integers. 
A: Follow-up
After much tinkering, I have settled on the following Peano-like axioms for the integers:
Let $Z, L, R, 0, s$ be such that:
$R\subset Z$, the non-negative integers (right)
$L\subset Z$, the non-positive integers (left)
$Z=R\cup L$
$\forall x (x\in R \wedge x\in L \leftrightarrow x=0)$
$s: Z\rightarrow Z$, a bijection 
$\forall x (x\in R \rightarrow s(x)\in R)$
$\forall x (x\in L \rightarrow s^{-1}(x)\in L)$
$\forall x (x\in R \rightarrow s(x)\neq 0)$
$\forall x (x\in L \rightarrow s^{-1}(x)\neq 0)$
$\forall P ((P\subset Z \wedge 0\in P\wedge \forall x (x\in P\rightarrow s(x)\in P) \wedge \forall x (x\in P\rightarrow s^{-1}(x)\in P) ) \rightarrow Z\subset P) $
A: This has a model in the naturals. Take $Z$ to be the naturals, $Pos$ to be the evens, $Neg$ to be the odds plus 0. Define $s$ to be $+2$ and $s'$ to be $+2$ except $s'0=1$.
A: It seems that you are attempting to describe a second order theory such that, up to isomorphism, all models are the integers. 
While your formulation is not entirely clear I think it can, with a bit more work, be turned into a finite list of second order axioms that will do the job. 
