Derive Peano Axioms from Properties of the Natural Numbers

Question

It is well-known that, beginning from Peano's axioms and the usual rules of logic, one can derive all the properties of the natural numbers, and beyond, up to the complex numbers. See Landau's Foundations of Analysis for the details.

What I am interested in is the reverse direction, and here it's not so much getting from the complexes to Peano. It would be sufficient for my purposes just to get from the properties of the natural numbers to the Peano axioms.

I have taken a look at this question, but as I am not an expert in foundations, this answer is beyond me (I'm looking for something on the level of Landau, actually). I'm also not sure it's really what I'm asking.

So, to be specific, I would like to know if, from this set of axioms (or a suitably altered set of axioms):

1. Closure: For all natural numbers $a$ and $b,$ $a+b$ is a natural number.
2. Commutativity: For all natural numbers $a$ and $b,$ $a+b=b+a$ and $ab=ba.$
3. Associativity: For all natural numbers $a,b,$ and $c,$ $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc).$
4. Identities: There exist natural numbers $0$ and $1$ such that $a+0=a$ and $a\cdot 1=a$ for all natural numbers $a.$
5. Distributive: For all natural numbers $a,b,$ and $c,$ $a(b+c)=ab+ac.$
6. No Nonzero Divisors: For all natural numbers $a,b,$ if $ab=0,$ then either $a=0$ or $b=0.$

we can derive the Peano axioms (slightly modified from Landau to include $0$):

P.1. $0$ is a natural number.

P.2. For each natural number $x,$ there exists exactly one number, called the successor of $x,$ which will be denoted by $x'.$

P.3. We always have $x'\not=0.$

P.4. If $x'=y',$ then $x=y.$

P.5. Let there be given a set $M$ of natural numbers with the following properties: a. $0$ belongs to $M.$ b. If $x$ belongs to $M,$ then so does $x'.$ Then $M$ contains all the natural numbers.

I'm also interested to know if we can weaken the first list of axioms above. Do we need closure? No non-zero divisors? Distributive law?

My Thoughts:

P.1 seems obvious, given Axiom 4. P.2, it seems to me, would need to be derived from the '$+1$' operation, which is well-defined given Axioms 1 and 4. I'm fairly sure, if you did that, P.4 would be fairly straight-forward. P.3 and P.5 seem difficult.

Many thanks for your time!