Can Peano's 9th axiom be expressed using a self-referential set definition? According to Wikipedia, the 9th axiom I am asking about is this.
I am wondering, is the following a valid way of expressing the same idea? If not, why not, and what would be the proper self-referential expression (if there is any)?
If $K = \{0\} \cup \{S(k)\mid k \in K\}$, then $K=\mathbb{N}$.
 A: Certainly this makes sense in some contexts.  Akiva Weinberger is reasonably concerned that such definitions can be circular and nonproductive, and gives the example $$S=\{x\mid x\in S\}.$$ But in your particular case there is no difficulty.  One just needs to be careful how the definition is formulated.
The idea is that one can define a mapping $$\mathcal M(X) = \{0\} \cup \{S(k)\mid k\in X\}.$$  That is, the mapping $\mathcal M$ isn't a set; it's a function that transforms one set into another.  For example, it transforms $\emptyset$ into $$\mathcal M(\emptyset) = \{0\} ,$$ and $$\mathcal M(\{2,7,23\}) = \{0, 3,8,24\}.$$  This is perfectly well-defined, not circular at all.
Now your proposed definition can be written:
$$
\text{If } \mathcal M(K)=K, \text{ then } K=\Bbb N.
$$
So one should investigate the fixed points of the mapping $\mathcal M$: that is the sets $F$ for which $\mathcal M(F) = F$.  It transpires that $\Bbb N$ is a fixed point of $\mathcal M,$ and that although it is not the only fixed point of $\mathcal M$, every fixed point does contain $\Bbb N$ as a subset.  (For example, $\Bbb Z$ is also a fixed point of $\mathcal M$.)
One can do even better: let $\mathcal F$ be the set of all fixed points of $\mathcal M$.  The foregoing shows that $\mathcal F$ has a smallest element, namely $$\bigcap F = \Bbb N.$$   This least fixed point is precisely the set we were trying to define in the first place.  The original definition, which appeared circular, turned out to be meaningful in a rather straightforward way.  We only had to formalize the idea of the mapping $\mathcal M$, and then show that $\mathcal M$ had a least fixed point.
This approach is more commonly used in the formal semantics of programming languages, to define the meaning of recursive functions.  Consider possible objections to your “circular” definition.  One could raise the same objections to the completely mundane computer program
factorial(n) :=
   if n = 0 then 1
            else n * factorial(n-1)

“You can't define the factorial function this way, it's circular,” someone might say.  But programmers do this all the time, it's perfectly fine, it compiles and runs and computes factorials, so the claim that it is “circular” is a bit silly, and just shows that there is something not quite obvious going on. The fixed-point explanation works here as well. The mathematical meaning of this definition is that there is a mapping $$\mathcal M(f) =  f' \text{ as follows}\\
\text{If } n = 0 \text{ then }\\ f'(n) = 1 \\
\text{else }\\ f'(n) = n \cdot f(n-1).$$
This is not circular all all; it defines $f' = \mathcal M(f) $ in terms of $f$ only.  But again one can ask about the fixed points of $\mathcal M$ and even the least fixed point. The least fixed point   (for the proper understanding of “least”)
is precisely the factorial function we were trying to define.
In a suitable programming language the same technique, and the same understanding, allows a recursive definition of $\Bbb N$ in the same way.  For example, in the real-world programming language Haskell, it is 100% correct to write
naturals = 0 : map (1 +) naturals

which is nearly an exact translation of your definition above: the sequence of natural numbers begins with zero, and follows with a sequence of every natural number increased by $1$.  Let me emphasize, this actually works in the sense that you can write this as part of a Haskell program that does something useful in the larger world of business or science. (For technical reasons I have defined $\Bbb N$ as a sequence rather than a set.)
One can prove that in a suitable semantic domain, every such definition has a least fixed point.  Taking up  Akiva Weinberger's objection, what is the least fixed point of the mapping $\mathcal W(S) = S$?  It turns out that the suitable semantic domain contains an element $\bot$, and the least fixed point of $\mathcal W$ is $\bot$, which computationally represents the (non-)result of an infinite loop.
A: Given a set $K$, the equation$$K=\{0\}\cup\{S(k)\mid k\in K\}$$ is either true or false. There is just one subset of $\Bbb N$ that satisfies that. Therefore, the statement

For any $K\subseteq\Bbb N$, if $K=\{0\}\cup\{S(k)\mid k\in K\}$, then $K=\Bbb N$

is perfectly fine and clear.
(I added the words "For any $K\subseteq\Bbb N$", but in the context of second-order Peano Arithmetic, all sets involved are subsets of $\Bbb N$, so it should be fine to omit this since it's clear from context.)
A: After some deliberation and useful inputs from MJD and Akiva Weinberger, I concluded that my formulation is not universally understood to be precise, and should be amended as follows:
$\mathbb{N}$ is defined as the least set $K^*$ within the poset of sets $K=\{0\}\cup\{S(k) \mid k \in K\}$ ordered by set inclusion.
See Wikipedia for the definition of poset and least.
