In what follows we shall assume the conventional interpretation of the symbols $'=','\ne'$ and the other logical symbols used to formulate definitions. Lowercase Latin letters $a,b,\dots,z$ will designate number variables. The notation $\mathcal{P}\left[n\right]$ represents a proposition regarding a number variable $n$. By a number we mean an element of the set $\mathbb{N}$ defined by the following axioms:
I. $1\in\mathbb{N}.$
II. $\underset{a}{\forall}\exists a^{\prime}\in\mathbb{N}.$
III. $a^{\prime}=b^{\prime}\implies a=b.$
IV. $\underset{a}{\forall}a^{\prime}\ne1.$
V. $\left(\mathcal{P}\left[1\right]\land\underset{n}{\forall}\left(\mathcal{P}\left[n\right]\implies\mathcal{P}\left[n^{\prime}\right]\right)\right)\implies\underset{n}{\forall}\mathcal{P}\left[n\right].$
The object $a^{\prime}$ is called the successor of $a$. That is my rendering of the axioms attributed to Peano.
We now define the set of functions
$$ \Phi=\left\{ \varphi_{a}:\mathbb{N}\to\mathbb{N}\backepsilon\varphi_{a}\left[x^{\prime}\right]\equiv\varphi_{a}\left[x\right]^{\prime}\right\} . $$
For notational convenience we also define
$$ \left[\_+\right]:\mathbb{N}_{1}\to\Phi $$
so that
$$ \left[a+\_\right]\equiv\varphi_{a}, $$
to be interpreted as $$ a+b:\mathbb{N}\times\mathbb{N}\to\mathbb{N}\text{ where }a+b\equiv\varphi_{a}\left[b\right]. $$
From the above definitions we may prove the associative and commutative properties of addition. By defining the symbols $'<'$ and $'>'$ as meaning
$$ a<b\implies\underset{c}{\exists}a+c=b\text{ and }a>b\equiv b<a, $$
we may define (or show) the well ordering of $\mathbb{N}.$
If we now define the notation $\bar{a}$ to mean $a<\bar{a},$ we have $$ \neg\underset{x}{\exists}x+\bar{a}=a. $$
That is, $\mathbb{N}$ contains no negative numbers. We also have the condition $$ \underset{a,b}{\forall}a+b\ne a. $$
Which says $0\notin\mathbb{N},$ or that there is no additive identity element in our set of numbers.
If we define numbers to be the set $\mathbb{N}_{0}$ by replacing the symbol $'1'$ with the symbol $'0'$, without bestowing additional properties to $'0'$, we arrive at exactly the same algebraic structure as we have for $\mathbb{N}.$
Let us assume we agree that by the natural numbers we intend the most fundamental complete set of numbers of interest to a mathematician. We shall further assume that the only reasonable choices are between $\mathbb{\mathbb{N}}$ and $\mathbb{N}_{0}$ as defined above. To argue that it is preferable to choose $\mathbb{N}_{0}$ rather than $\mathbb{N}$ presupposes that the element which is not a successor has a different interpretation when symbolized as $'1'$ than when symbolized $'0'.$
So my question is this: is the basic algebraic structure intended by $\mathbb{\mathbb{N}}=\left\{ 1,2,\dots,0\right\} $ distinct form that intended by $\mathbb{N}_{0}=\left\{ 0,1,2,\dots\right\} ?$
Clearly, $n\mapsto n+1$ is an isomorphism between these two, hence both have the same logical structure. If you prefer to always work with $\Bbb N$, you may want to introduce he operation $+^*$, where $a+^* b$ is the unique $c$ with $c'=a+b$. Then $+^*$ is associative, abeliam amd does have a neutral element - an operation that sometimes comes in handy. If you need suc an operation on $\Bbb N$, you can use $+^*$ - or work in the isomorphic $\Bbb N_0$, where $+^*$ becomes ordinary addition. This makes $(\Bbb N,+^*)$ isomorphic to $(\Bbb N_0,+)$, but of course $(\Bbb N,+)$ and $(\Bbb N_0,+)$ are not isomorphic.