An arithmetic function on $N$ can be seen as a subset of $N^3$. For a function $f$ of two variables on $N$ (like addition or multiplication),
$\forall a,b,c ((a,b,c)\in f\rightarrow (a,b,c)\in N^3)$
$\forall a,b\in N\exists c\in N ((a,b,c)\in f)$
$\forall a,b,c,d\in N ((a,b,c)\in f\wedge (a,b,d)\in f\rightarrow c=d)$
If you want to construct the add function on $N$ using only the successor function $s$, you start by selecting a subset $S$ from $\mathcal P(N^3)$ such that:
$\forall a(a\in S \leftrightarrow a\in \mathcal P(N^3) \wedge \forall b\in N((b,1,s(b))\in a)\wedge\forall b,c,d\in N ((b,c,d)\in a\rightarrow(b,s(c),s(d))\in a))$
Then the required add function (a subset of $N^3$) is just the intersection $\bigcap S$.
UPDATE
Alternatively, we can construct the set of ordered triples $add$ as follows:
$\forall x,y,z:[(x,y,z)\in add \iff (x,y,z)\in N^3$
$\land \forall a\subset \mathcal P(N^3):[\forall b\in N:[(b,0,b)\in a]\land \forall b,c,d:[(b,c,d)\in a \implies (b,s(c),s(d))\in a]$
$\implies (x,y,z)\in a]] $
Then you can prove that $add$ is a function such that:
$\forall x\in N: add(x,0)=x$
$\forall x,y \in N: add(x,s(y))=s(add(x,y))$