Edit: I've edited this question quite a bit since I first posted it, with the intention of making it clearer what I am asking. If something still seems unclear, please let me know. Check the edit history for previous versions.
Motivation
When I ask for the definition of the natural numbers, the only consistent response which I receive is that "the natural numbers are the set containing < 0 / 1 > which is closed under succession" (from this point on, I will consider $0$ to be the smallest natural number). All additional information about the natural numbers seems to vary depending on the source.
Since this is the common point of agreement, I can conclude that at least the following are true of the natural numbers:
$1.\qquad0 \in \Bbb{N}$
$2.\qquad\forall n\in \Bbb{N}. s(n)\in \Bbb{N}$
This alone does not suffice to define the natural numbers, however. It is trivial to show that if $s(n)=n$, then $\{0\}=\Bbb{N}$ satisfies (1) and (2).
So there must be an additional rule $\forall n\in \Bbb{N}.s(n)\ne n$.
This is still insufficient to define the natural numbers. If $s(0)=1$ and $s(1)=0$, then $\{0,1\}=\Bbb{N}$ satisfies all three rules.
After doing this about six more times, I came up with a list of rules which, taken together, seem to identify the set of natural numbers.
Upon inspection, I noticed that these same rules can be used to define other things aside from "the set of natural numbers", such as ordinals (or at least countable ordinals), polynomials, and 'small' cardinals (e.g. $\aleph_0,\aleph_1,\ldots$).
To that end, I elected to abstract the rules so that they could be inherited by these other systems.
The Rules
Let $(X,\sigma,\prec)$ be a structure satisfying the following:
$1.\qquad \exists o.o\in X$
$2.\qquad \forall x\in X.\sigma(x)\in X$
$3. \qquad \forall x\in X. x\prec\sigma(x)$
$4. \qquad \forall x,y \in X. x\prec y\implies x\ne y$
$5. \qquad \forall x,y,z \in X. x\prec y \land y\prec z\implies x\prec z$
Note: $(X,\sigma,\prec,o)$ may be used instead if you wish to discard $(1)$
Note: The natural numbers are obtained by $X\to\Bbb{N}$, $\sigma\to s$, and $\prec\to<$.
Were I just a bit more demanding, I would say that these rules are inadequate because they fail to specify the full extent of $X$. However, I never saw this as a possibility for the motivating example (the natural numbers), and seeing as what I do have is more precise than anything I have been given prior, this seems like a small enough concession.
Ideally, $X$ can be obtained by taking $\{o\}$, computing $\sigma(o)$, adding this to $\{o\}$ and continuing in this fashion for an indeterminate number of stages.
(technically, this would also require $\forall x\in X\setminus \{o\}. \exists w\in X.\sigma(w)=x$ and $\nexists x\in X.\sigma(x)=o$, but I omitted these since they are not essential to the question).
This could all be summarized as $X=\{o,\sigma(o),\sigma(\sigma(o)),\ldots\}$ - and it's tempting to leave it at that. But what does "$o,\sigma(o),\sigma(\sigma(o)),\ldots$" actually mean? Does it terminate at some point? What is the eventual cardinality of $X$?
Rather than casually dismissing these questions with the usual "we know the natural numbers are countable [because reasons], and therefore..." I resolved to find a direct and definitive answer.
Since it isn't possible to continue the sequence $\sigma(o),\sigma(\sigma(o)),\ldots$ forever I decided to use ordinals to define the iteration of the function $\sigma$:
$\sigma^0(x)=x\qquad (0\in\mathbf{Ord})$
$\sigma^{s(\alpha)}(x)=\sigma(\sigma^\alpha(x))\qquad (\alpha\in\mathbf{Ord})$
Why ordinals?
Clearly there is no greatest member of $X$, so $X$ is something to the effect of "all elements up to $\sigma^\infty(o)$". But the notation $\sigma^\infty(o)$ is ambiguous. What is $\sigma^\infty(x)$? What is $\sigma(\sigma^\infty(x))$? Using ordinals, it is possible to differentiate infinite iterations of $\sigma$ from one another in a way that is impossible if we just leave it at "$X$ is infinite". Since $(X,\sigma)\cong(\Omega,s):\Omega\subseteq\textbf{Ord}$, the repeated applications of $\sigma$ correspond directly to successive ordinals.
The Question
Originally, this question was about restricting the iteration $\sigma$ to clarify the definition of $X$. At the time, I had assumed that "closed under $\sigma$" was an absolute - i.e. "closed under arbitrary applications of $\sigma$" by definition - and that a restriction to the iteration of $\sigma$ was necessary for $X$ to remain countable. The need to explicitly state the restriction on the repeated application of $\sigma$ might seem unnecessary at first, but it is no more necessary to say that a topology (using the open-sets definition) is "closed under finite intersections" as opposed to "closed under intersection".
Regardless, unless $X$ is a proper class (in which case, the natural numbers are also a proper class), there must be some least ordinal $\beta$ such that $X=\bigcup_{\alpha<\beta}\{\sigma^\alpha(o)$}.
This is a problem because $\beta$ cannot be determined via 1 through 5. Interpreting $X$ as the set of natural numbers with succession and their usual order, what this effectively says is that "the natural numbers are the set containing $0$ and all successors of $0$ which are natural numbers" - which is of course true, but it's also unhelpful.
Since Tanner Swett's answer suggests that 1 through 5 can be used to prove that $X$ is closed only under finite iterations of $\sigma$, how can I show this to be the case?
Furthermore, how do I prove that $\beta=\omega$ (which it must, if $X$ is deemed countable)?
Finally, how can I define $X$ in a non-circular fashion? Even if I can prove that $\omega$ is the least ordinal such that $X=\bigcup_{\alpha<\omega}\sigma^\alpha(o)$, I still haven't defined $\omega$. And since $(\omega,s,\in)$ is an example of the structure being described, this ultimately reduces to $\omega=\bigcup_{\alpha<\omega}s^\omega(\emptyset)$ - which, along with 1 through 5, is trivially satisfied by every infinite ordinal (again, true but unhelpful).
Note: applying the restriction "for finitely many iterations of $\sigma$" doesn't seem to work using the Cantor or Dedekind definitions of finite, since applying this construction to the class of Cantor finite sets leads back to the same problem and using the Dedekind definition makes things... weird. I'm still working on the Kuratowski definition.
Note: I have not studied model theory in-depth, but I tagged this as "model theory" since that seems to be the subject which came up most frequently while researching this question.
Note: while I am calling this a "construction" in the title, I would prefer to think of it as a "definition," since a failure to equate the two, in this case, would entail that the natural numbers are not defined - which is something I am unwilling to accept.
I'm not sure what you mean by "valid," exactly, but it's not accurate to say that if $\varphi$ is a predicate on $X$, then $\forall x_0\in X.I_{x=x_0}^{\varphi}\sigma(x)\in X$. This is because—for one—your definition of $I$ is not complete. For example, if $\phi(x) = (x = x)$, then $I_{x=o}^\phi\ \sigma(x)$ is an undefined expression.
Well, that's not true at all. If $\gamma = \omega_1$, then $I_{n = \emptyset}^{i \in \gamma}\ s(n)$ is an undefined expression.
And besides, (2) holds of $X = \omega$, and yet $\omega_1 \notin \omega$.
Now, you write in a comment:
Well, technically, "closed under $\sigma$" means "closed under just one application of $\sigma$," which entails being closed under any finite number of applications of $\sigma$. But it doesn't entail being closed under infinitely many applications of $\sigma$ (that's not even a well-defined concept in general).
So, there's no problem here. The set of natural numbers is closed under succession, and it satisfies your definition of a structure $X$. This doesn't lead to any falsehoods such as $\omega_1 \in \omega$.
It looks like what you've done is to assume that the conditional iteral $I$ is well-defined and well-behaved everywhere, and from this, you've derived a falsehood. Thus, you've proven that the conditional iteral $I$ is not well-defined and well-behaved everywhere.
You write:
You haven't defined $I$ in such a way that $I_{n=\emptyset}^{i\in\omega_1}s(n)$ is defined as $\omega_1$... well, not unless the word "hypothetical" in your definition is meant to imply that $I$ is defined even if your program does not halt. But in that case, it's not clear what $I$ is defined as. Saying something like "it's the hypothetical output value of this program which runs forever without outputting a value" doesn't constitute a meaningful definition.
(Side question: what does the $i$ in the superscript in $I_{n=\emptyset}^{i\in\omega_1}s(n)$ mean? Is that supposed to be $n$?)
In any case, axiom (2) states that $\forall x\in X.\sigma(x)\in X$; it doesn't say anything about $I_{n=\emptyset}^{i\in\omega_1}s(n)$. You cannot conclude, from the statement that $\forall x\in X.\sigma(x)\in X$, that $I_{n=\emptyset}^{i\in\omega_1}s(n) \in X$.
Overall, you seem to think that the first-order logic sentence $\forall x\in X.\sigma(x)\in X$ entails that you can take an element of $x$, apply the function $\sigma$ infinitely many times, and end up with an element of $x$ as the result. It does not.