Generalizations of PA and its standard and non-standard models Consider Peano's axioms — in its first-order version and without addition and multiplication — with its single injective function $S$:


*

*$(\forall x) \neg Sx = 0$

*$\Big(\phi(0)\ \ \&\ \big( (\forall x)\ \phi(x) \Rightarrow \phi(Sx)\big)\Big) \Rightarrow (\forall x)\ \phi(x)\ \ \ $ for all first-order formulas $\phi(x)$


These axioms fix $\mathbb{N}$ as its standard model and $\mathbb{N} + \mathbb{Z}$ as its standard non-standard model.
Now consider a generalization with finitely many pairwise injective functions $S_i$, $i = 1,...,n > 1$:


*

*$(\forall x) \neg S_ix = 0\ \ \ $ for $i = 1,...,n$

*$\Big(\phi(0)\ \ \&\ \big( (\forall x)\ \phi(x) \Rightarrow (\phi(S_1x)\ \&\ ...\ \& \ \phi(S_nx))\big)\Big) \Rightarrow (\forall x)\ \phi(x)\ \ \ $ for all formulas $\phi(x)$


The standard model of these axioms is a rooted, directed, and appropriately edge-colored tree. 

Question 1: But how does a non-standard model look like, containing a directed
  tree, infinite to-and-fro (as a generalization of
  $\mathbb{Z}$)? What is an example of such a tree?

Finally consider a generalization with countably many pairwise injective functions $S_i, i \in \mathbb{N}$. It's (relatively) easy to imagine an appropriate rooted tree $T_\omega$ — with $\omega$-many childs of each node — as a standard model.

Question 2: But of what first-order theory, actually?

Axiom (scheme) 1 is not a problem, but can the induction axiom (scheme) 2 be generalized for infinitely many functions?
While $\mathbb{N}$ can be at least partially captured — as a standard model — by a first-order theory, the corresponding $T_\omega$ cannot be captured at all by a first-order theory?
 A: Let me begin by describing a particular non-standard model $M$ of the theory of $n$ successor functions, i.e., of the complete first-order theory of the $n$-fold branching tree $T$.  $M$ consists of $T$ plus the set $A$ of all infinite sequences of elements of $\{1,2,\dots,n\}$.  $S_i$ acts on $A$ by sending any sequence $(a_0,a_1,\dots)$ to $(i,a_0,a_1,\dots)$.  
One might call $M$ "the standard non-standard model", but the analogy with $\mathbb N+\mathbb Z$ fails in at least one respect: $M$ has many nonstandard components. In more detail, consider the graph whose vertices are the elements of the model and whose edges connect any element to any of its $n$ successors.  The graph corresponding to the standard model is connected.  When $n=1$, the graph corresponding to $\mathbb N+\mathbb Z$  has just two components.  But when $n>1$, the graph corresponding to my $M$ has $\mathfrak c$ (the cardinal of the continuum) connected components.  Two elements of $A$ are in the same component iff they become equal after you delete some finite initial segments (possibly of different lengths) from each of them.  If an analog of $\mathbb N+\mathbb Z$  should have just two components, then it would have to consist of $T$ plus a single component of $A$, and I see no good reason to prefer one component over another.
A general model of the complete first-order theory of $T$ would consist of $T$ plus, for each component of $A$, some cardinal number of disjoint copies of that component.  
Thus, for $n>1$, this first-order theory is not $\omega$-stable (and in particular not $\aleph_1$-categorical, in contrast to the $n=1$ case), but, unless I'm overlooking something, it is superstable.  (In particular, whether or not $n>1$, it does not have a definable linear ordering, contrary to what some comments seem to have presupposed.)
A: Regarding Question 2, rather than having finitely many function symbols $S_n(x)$, the easiest solution is to have a single relation symbol $S(n,x,m)$ asserting that the "$n$-labeled" successor of $x$ is $m$.  Then your induction axiom can quantify over this:
$$ \phi(0)\ \&\ {\Large[}(\forall x,n,m)\ \phi(x)\ \&\ S(n,x,m)\ \Rightarrow\ \phi(m){\Large]}\ \Rightarrow\ (\forall x)\phi(x)$$
The only problem remaining is that the scheme above allows the labels themselves to be "tree-shaped natural numbers".  If you want the labels on the edges to be "linear natural numbers" (i.e. only one successor for each element), you'll need to include a separate copy of Peano's axioms in your theory, and simulate a two-sorted theory by including a unary relation (predicate) symbol which distinguishes "linear" naturals from "tree-shaped" naturals and asserts that an element of one kind is never the successor of an element of the other kind.
A: There are apparently missing axioms in the question. “Peano arithmetic” in the signature $\{0,S\}$ is not axiomatized by the two axioms given, since the axioms have a two-element model $\{0,\infty\}$ where $S(0)=S(\infty)=\infty$. One needs further axiom
$$\forall x\,\forall y\,(S(x)=S(y)\to x=y)$$
(expressing that $S$ is injective).
Now, the “standard model” with injective functions $\{S_i:i\in I\}$ (where $I$ can be finite or infinite of arbitrary cardinality) is nothing else than the term algebra (with no variables) in the signature $\{0\}\cup\{S_i:i\in I\}$. The complete theory of term algebras is well known, it can be easily axiomatized, and it enjoys quantifier elimination in an appropriate language (see e.g. Hodges’ “A shorter model theory” for details). In our case, the theory is axiomatized by


*

*$\forall x\:S_i(x)\ne0$ for each $i\in I$

*$\forall x\,\forall y\:S_i(x)\ne S_j(y)$ for each $i,j\in I$, $i\ne j$

*$\forall x\,\forall y\,(S_i(x)=S_i(y)\to x=y)$ for each $i\in I$

*$\forall x\:S_{i_0}(S_{i_1}(\cdots(S_{i_n}(x))\cdots))\ne x$ for each $n\in\omega$ and $i_0,\dots,i_n\in I$

*$\displaystyle\forall x\,\Bigl(x=0\lor\exists y\,\bigvee_{i\in I}S_i(y)=x\Bigr)$ if $I$ is finite


In particular, in the finite case the induction schema follows from these axioms. On the other hand, the axioms 2, 3, and 4 do not follow from Hans’ axioms; in the case of 3 (the injectivity axiom), this is likely an omission (since the functions are described as injective), but I’m not sure about the others (though I find it unlikely that intended nonstandard models include the natural numbers endowed with $S_1(x)=\dots=S_n(x)=x+1$, or the model $\{0,a_n,b_n:n\in\omega\}$ where $S_1(0)=a_0$, $S_2(0)=b_0$, $S_1(a_{2n})=b_{2n}$, $S_2(a_{2n})=a_{2n+1}$, $S_1(b_{2n})=b_{2n+1}$, $S_2(b_{2n})=a_{2n}$, $S_1(a_{2n+1})=a_{2n+2}$, $S_2(a_{2n+1})=b_{2n+1}$, $S_1(b_{2n+1})=a_{2n+1}$, $S_2(b_{2n+1})=b_{2n+2}$). Notice also that in the Peano case $|I|=1$, axiom 4 follows from axioms 1, 3 and induction, but this is no longer true when $|I|\ge2$.
This axiomatization answers the original Question 2. Moreover, it may be helpful for understanding the structure of nonstandard models of the theory (since the induction-free axioms are more concrete) and to correct the axiomatization (if I’m right in believing that the incompleteness of Hans’ axioms is unintended).
