Part 1
You can't necessarily assume that these rogue numbers exist, however, under the first four axioms, you can't say definitively that they don't exist, which is really the point.
Suppose that in addition to the expected natural numbers $0,1,2,$ etc., there are numbers $a,b,$ and $c$ so that $a{+\!+}=b,\ b{+\!+}=c,$ and $c{+\!+}=a$. This does not contradict any of the first four axioms, so it is a valid construction under those axioms. However, it does not correspond to what we expect of the natural numbers, so there is another axiom to restrict this construction.
Part 2
The property $P(n)$ is arbitrary, it does not need to be specifically defined for the purposes of the axiom. Informally, the fifth axiom states that induction works. Given a property $P(n)$, the two statements that $P(0)$ is true and if $P(n)$ is true, then $P(n{+\!+})$ is true are sufficient to show that $P(n)$ is true for all natural numbers $n$.
Put another way, this means that $0$ and all of its iterated successors make up all of the natural numbers. In the previous example, with $a,b,$ and $c$, there is no iterated successor of $0$, that is, no "regular" natural number $n$, where $n{+\!+}=a$. So, this construction is no longer valid with the axiom of induction.
Edit:
I chose a finite set of hypothetical numbers in Part 1 simply for convenience. What properties these objects have, and what their successors are, is irrelevant. All that matters is that they fit the first four axioms, in that everything in the set has a successor that is also in the set, nothing has a successor that is $0$, and no two distinct objects have the same successor. There are, in fact, infinitely many things that could be in this set, and no way to define them all. In addition to the problem of not being numbers that we "want," this also means that the natural numbers, as defined so far, are not uniquely defined up to isomorphism.
There is one other property that we know these rogue numbers have, and that is whatever we use to define them as "rogue". That is, while they are a valid part of the natural numbers that we have constructed so far, they aren't a part of the natural numbers that we want to construct. So the natural question (no pun intended) is what other property or properties do we want the natural numbers to have?
Naively, the natural numbers are the "counting" numbers. For any natural number $n$, if we start at zero and repeatedly take the successor, we would (theoretically) be able to get $n$, eventually. So, presumably, the rogue numbers are numbers that you can't "count" to. This is what the fifth axiom, the axiom of induction, deals with. If $P(0)$, and $P(n)\Rightarrow P(n+1)$, then you can show $P(1)$, $P(2)$, and so on for any number that you can count to from zero. The axiom states that this covers all of the natural numbers. Everything else is excluded. It doesn't matter if they are irrational numbers, Gaussian integers, letters, transfinite and infinitesimal surreals, or, as Kundor commented, wooly mammoths. If you can't count to it from zero, it isn't a natural number.
Use induction on the statement "$n=0$ or there exists a natural number $b$ such that $b++=n$."
Showing this for $n=0$ is obvious. Do you need help on showing the induction step?
ADDED: Here is the induction step (in summary: you can fill in the small gaps).
Assume the statement is true for $n$. There are two cases: $n=0$ or $n \ne 0$.
If $n=0$, then $0++=n++$, and the statement is true for $n++$.
If $n\ne 0$, we must have $b++=n$ for some $b$ (since we assumed the statement is true for $n$). Then $(b++)++=n++$, and again the statement is true for $n++$.
ADDED EVEN LATER:
@Andreas Blass gave a much shorter and better induction step in his comment. Here it is:
Assume the statement is true for $n$ (although we will not use this assumption). Then let $b=n$. We trivially get $b++=n++$, so the statement is true for $n++$.
Best Answer
The proof is by induction on $a$.
Basis [case with $a=0$]. We want to prove that :
But we have that : $0+b=b$, by definition of addition.
And also : $0+c=c$.
Thus, by transitivity of equality : $b=c$.
Induction step. Assume that the property holds for $a$ and prove it for $a'$ [I prefer to use $a'$ instead of $a++$ for the successor of $a$].
This means :
We have $a'+b=(a+b)'$ by definition of addition.
And $a'+c=(a+c)’$, by definition of addition.
We assume that : $a'+b=a'+c$, and thus, again by transitivity of equality, we have that : $(a+b)'=(a+c)'$.
By axiom 2.4 (different numbers have different successors), by contraposition, we conclude that $(a+b)=(a+c)$.
Thus, applying induction hypothesis, we have that :
The general strategy used above in order to prove "if $P$, then $Q$", is to assume $P$ and derive $Q$.
This type of proof is called Conditional Proof; we can see it above :
1) if $a+b=a+c$, then $b=c$ --- induction hypothesis
2) $a'+b=a'+c$ --- assumption for CP
3) $(a+b)'=(a+c)'$ --- from 2) and definition of addition and tarnsitivity of equality
4) $a+b=a+c$ --- from 3) and axiom 2.4 by Contraposition [the axiom says : "if $(a+b \ne a+c)$, then $(a+b)' \ne (a+c)'$"; thus, contraposing it we get : "if $(a+b)' = (a+c)'$, then $(a+b) = (a+c)$"] and Modus ponens
5) $b=c$ --- from 4) and 1) by Modus ponens (also called : detachment)