Comment by im3w1l
It's interesting to consider the possibilities that you have if the recursion axiom is removed. Call those numbers the super-natural numbers. Let's state the recursion axiom A, and remove it.
A: If K is a set such that: 0 is in K, and for every supernatural number n, n being in K implies that S(n) is in K, then K contains every super-natural number.
Without A, there may be a set S that contains 0 every successor of 0 (that is it contains all the natural numbers), but still does not contain every super-natural number. There are three possibilities: There may some copies of N (e.g. a 0' successors 1', 2' etc). There may be some copies of Z (there is an axiom that no number has 0 as a successor, but 0' could be preceeded by -1'). And there may be some copies of Z_k (e.g. 0' followed by 1' followed by 2' followed by 0' again). We could call every copy of N, Z or Z_k a "branch" of the supernaturals.
I say may, because the axioms leave it open, it could exist or could not exist.
Now what happens if you define addition with the supernatural numbers? Let a and b be supernatural numbers. We will use the regular definition
a+0 = a
a+S(b) = S(a+b)
What happpens when we do this?
First let n be a natural number. Due to the recursive property of natural numbers a+n=S^n(a). So we can add a supernatural number on the left side to a natural number on the right side just fine. But what if we have a proper supernatural number on the right side? The definition is completely silent on the matter. But could we find a valid extension?
Thinking about it a little bit I found that defining a+b=b (for any b that is not a natural number) will be a consistent (but perhaps not very interesting or useful) extension.
Notice that this definition will pick the branch of the result based on b, except in the case where b is natural. This is the not a coincidence, for if b is from a Z_k branch then the result must also be from a Z_k branch, as can be found by applying the successor k-times (it could I suppose in theory be from a different Z_k branch though).
E.g.
If b is from a Z_3 branch then b=S(S(S(b))) meaning a+b=a+S(S(S(b)))=S(a+S(S(b)))=S(S(a+S(b)))=S(S(S(a+b))) so a+b is also from a Z_3 branch.
I think the traditional word for the "super-natural number" is "nonstandard integer".
You correctly notice that in case of nonstandard integer, the recursive definition alone is ambiguous, because while the standard integers are connected to zero by a finite chain of successor operations, the nonstandard integers are only chained to each other in infinite chains unconnected to zero. So you could have multiple implementations of a recursive function, each of them giving the same value for the standard integers, but different values for the nonstandard ones.
But there is one extra constraint that I think you didn't take into account. Peano axioms contain the "axiom of induction", which... if you look at it from a certain perspective, says: "whatever (first-order statement) is true for standard integers, it must also be true for nonstandard integers". Well, it doesn't say that directly; it's more like "whatever is true/false for some integers, there must be a smallest integer for which it is true/false".
This further constrains the possibilities for the "+" operation. If you can e.g. prove for the standard integers that "a+b = b+a", then according to this axiom, the same must also be true for nonstandard integers. So if "nonstandard + standard" is defined unambiguously, then so is also "standard + nonstandard".
But this still leaves some space for ambiguity in defining "nonstandard + nonstandard".