We are to show $\forall m,n,p\in \mathbb{N_{0}}:m+p=n+p\implies m=n$.
Proof. Let $m,n\in \mathbb{N_{0}}$.
Base case: If $p=0$, then $m+p=m$ and $n+p=n$ which clearly establishes $p=0\implies (m+p=n+p\implies m=n)$
Induction step: For a fixed $q\in \mathbb{N_{0}}$ we are to prove that \begin{align*}(m+q=n+q\implies m=n)\implies \Big(m+(q+1)=n+(q+1)\implies m=n\Big)\tag{$i$}\end{align*} We proceed by way of contradiction and to that end we suppose there exist $q\in \mathbb{N_{0}}$ such that $\sim (i)$. Since we know $q$ exists we may let $q=1$ so from $(i)$ we know \begin{align*}m+1=n+1\implies m=n\tag{$ii$}\end{align*} and \begin{align*}(m+1)+1=(n+1)+1\tag{$iii$}\end{align*} Since $m$ and $n$ in $(ii)$ can by any numbers and because $(m+1)$ and $(n+1)$ are numbers themselves, from $(iii)$ by applying $(ii)$ we know $m+1=n+1$ and, again, by $(ii)$ we've showed $m=n$ which contradicts our assumption $\sim (i)$.
Is it wrong this attempt of proof? If it is, may you help me by providing a proof by contradiction of this basic statement?
Addition of natural numbers have been defined recursively as- $$a+0=a$$ $$a+S(b)=S(a+b)$$ where $S$ is the successor function.
So, without assuming axiom of successor, you cannot even define addition on $\mathbb N$. So, there's no question of a cancellation law.