Axiomatic definition of integers
It's the unique commutative ordered ring whose positive elements are well-ordered.
The ring $\mathbb{Z}$ is the unique ordered ring which satisfies full second-order induction: $$\forall X(0 \in X \land (\forall n \geq 0)(n \in X \to n+1 \in X) \to (\forall n \geq 0)(n \in X)),$$ where $X$ varies over all subsets of $\mathbb{Z}$ (or even all sets). In the comments, Martin Brandenburg has given yet another characterization of $\mathbb{Z}$ which does not assume the ordering.
A dual characterization is that every nonempty subset of $\mathbb{Z}$ which is bounded below has a minimal element. This is closer to the characterization of $\mathbb{R}$. Note that all of these characterizations only make sense in standard second-order logic, but the proposed characterization of $\mathbb{R}$ has the same problem.
The ring of integers also has categorical characterizations. For example, as proposed in the comments, $\mathbb{Z}$ is initial object in the category of (ordered) rings. See this question for related information.
Part 1: Abelian group introduction
(See Part 2 for the final answer).
I feel that for a fully elegant definition of integers one needs to readdress the definition of an Abelian group. For the sake of communication I will even introduce a synonym minusop for abelian groups to stress the independence of their definition from the general groups. The general groups have an elegant definition (perhaps more than one). Then adding the commutativity axiom one gets Abelian groups. This seems to me unnecessarily complex. Let me define Abelian groups directly.
Symmetric operations
Symmetric operations should not be confused with much more special commutative operations.
By definition, an operation $\#:X^2\rightarrow X$ is called symmetric $\Leftarrow:\Rightarrow$
$$ \forall_{x\ y\ z\in X}\qquad x\ \#\ (y\ \#\ z)\ \ =\ \ z\ \#\ (y\ \#\ x) $$
This property will appear in the definition of minusop as an axiom. Both addition and subtraction are symmetric operations in any Abelian group. Every commutative operation is symmetric.
Minusop
By definition, a minusop is an ordered pair $(X\ -)$, where $X$ is an arbitrary set, and $-$ is a binary operation in $X$, such that the following three axioms hold:
- $x-(y-z)\ =\ z-(y-x)$
- $x-x\ =\ y-y$
- $x-(x-x)\ =\ x$
for arbitrary $x\ y\ z\in X$.
Every Abelian group admits a standard interpretation as a minusop; and every non-empty minusop admits its standard interpretation as an Abelian group, so that Abelian groups and non-empty minusops are essentially the same objects.
(I am not writing the last obvious statement in any detail to keep this post sensibly short).
REMARK
A class of operations even more general than symmetric is still useful--I call operation $\#:X^2\rightarrow X$ insider trading (in mathematics it's legal) $\Leftarrow:\Rightarrow$
$$\forall_{u\ w\ x\ y\ \in\ X}\qquad (u\ \#\ w)\ \#\ (x\ \#\ y)\ \ =\ \ (u\ \#\ x)\ \#\ (w\ \#\ y)$$
Every symmetric operation is an insider trading.
A simplification by Emil Jeřábek
The last two of the three minusop axioms above (in the previous section before REMARK) can be replaced by one, as pointed out Emil Jeřábek, as now presented below. A minusop can be axiomatized by the following two conditions:
- Symmetry: $x-(y-z)\ =\ z-(y-x)$
- Zero: $x-(y-y)\ =\ x$
for arbitrary $x\ y\ z\ \in\ X$.
It can be seen instantly that the earlier three axioms imply the two above (the new first axiom is a repetition of the old first axiom).
In the other direction, a substitution of $y$ by $x$ in axiom zero gives the old axiom 3. Furthermore, assuming the above two axioms we obtain:
$$ x-x\ =\ (x-x)-((y-y)-(y-y))\ =\ (y-y)-((y-y)-(x-x)) $$ $$ (y-y) - (y-y)\ =\ y-y $$
which proves the old axiom 2. The new 2-axiom system is equivalent to the old 3-axiom system.