How do Peano Axioms imply "nextness" with the successor? Going with this explanation of Peano's Axioms, I cannot understand how/where the successor function is definitively stated to be the very next number in the case of natural numbers. In this treatment, it says 


  The successor of $x$ is sometimes denoted $S⁢x$ instead of $x′$. We
  then have $1=S⁢0$, $2=S⁢1=S⁢S⁢0$, and so on.


Again, I don't see from the axioms how very next is guaranteed. Neither further on when addition is defined "based on the axioms." It seems a successor function could map "nextness" in many ways, e.g. $Sx$ could be $x+56$, or whatever as long as $Sx$ was something "further up the natural number line." So yes, I know I'm missing something here. . . .
If you just restrict yourself to the first axioms listed on that page, then indeed you could say that the expressions $0,S(0), S(S(0)), ...$ denote $0,2,1,4,3,6,5,....$ respectively. That would still satisfy the axioms. But you would still have a structure that is isomorphic to the natural numbers.

Also, if you define: 

$$\forall x \forall y (x &lt; y \leftrightarrow (y=S(x) \lor \exists z (y = S(z) \land x &lt; z))$$ 

(which I would think captures the notion of $&lt;$), 

then using only those very first axioms, you can prove that: 

$$\forall x \neg \exists y (x &lt; y \land y &lt; S(x))$$

i.e. that there is no number between any number and its successor ... and thus that the successor indeed gives you the very next number.

Finally, if you use the addition and multiplication axioms stated later on on that website, then you can prove some results that are certainly very suggestive of $S(0)$ working in a way consistent with our concept of the number $1$.

For example, it follows from the Peano Axioms that $$\forall x ( x · S(0) = x)$$ .... which would make sense if $S(0)$ would take the role we normally reserve for $1$ rather than, say, $56$

Likewise, you can prove from the Peano Axioms that $$\forall x ( x · S(S(0))= x + x)$$ ... again, that would square with the way we think about numbers if $S(S(0))$ is seen as $2$, rather than $117$.
First, I am not alone in asking this question of "nextness." The following is from the Wikipedia treatment, Peano Axioms:


  However, considering the notion of natural numbers as being defined by
  these axioms, axioms 1, 6, 7, 8 do not imply that the successor
  function generates all the natural numbers different from 0. Put
  differently, they do not guarantee that every natural number other
  than zero must succeed some other natural number.


...which goes on to use the often included axiom of induction to rework the successor function into implying nextness as always "one more." But then the article describes addition as defined recursively by two phantom Peanos:

\begin{align}
    a + 0 &amp; = a \\
 a + S(b) &amp; = S(a + b)
\end{align}

If the second expression is considered an axiom, then yes, $S$ would have to be consistently a "one more" successor. That is, there cannot be an $S(b) + a$ that is different from an $S(c)$ where $c = a+b$. However, these addition axioms are not really original to the PAs.

We might fall back on insisting that a successor function cannot from one application to the next behave differently, i.e., (allow me to switch successor notation) $m \succ m'$ implying $m'-m \geq 1$, and then $n \succ n'$ yielding $(n'-n) \neq (m'-m)$ is a violation of the spirit of the successor function and its brother induction.

It is true that $m=n \iff m'=n'$ implies uniqueness of the successor function only in the sense of injection (as pointed out in a comment by Somos). We might say a successor function must be an injective function producing distinct results, e.g.,  $n'$ is distinct from the successors which produced $1,...,n$, but I'm still doubtful this implies "one more," i.e., $n'$ to $n''$ proceeds one "very next thing" in a strict order at a time.

However, we might be rescued by Kleene's (Introduction to Metamathematics; p.20) example where he leverages the PA that $x' \neq 0$, or, $0$ is not the successor to any natural number. We might set up a contradiction: $$0''''=0''$$ but this would be true only if $$0'''=0'$$ Applying this logic again, $$0'''=0'$$ only if $$0''=0$$ So yes, this is a contradiction; hence, we have butted up against the lower boundary of the successor concept, which, therefore, implies a successor being greater than since we can run any successor-based induction of $\mathbb{N}$ backwards to $0$. Ironically, this is a sort of reverse-induction to prove that the Peano successor is consistently "one more" demanding some sort of $\mathbb{N}$-like order.

All in all, I'm still stuck on this....