In Tao's Real Analysis he defines addition:
Let $m$ be a natural number. To add zero to $m$, we define $0+m \equiv m$. Now suppose inductively that we have define how to add $n$ to $m$. Then we can add $n++$ to $m$ by defining $(n++) + m \equiv (n+m)++$.
Later he states:
Right now we only have two facts about addition: that $0+m=m$, and that $(n++)+m = (n+m)++$.
Do we not also get from this simple addition? $n + m = z$ for example.
If I understand correctly, the answer is yes: These axioms allow you to calculate $m+n$ for every pair of natural numbers $m, n$ (viewing $m, n$ as expressions of the form "$(...(0++)++)++)...)$").
To see this, consider the following example: $$2+2=[1+2]++=([0+2]++)++=[2++]++=(((0++)++)++)++.$$ (Here I use e.g. "$2$" as shorthand for "$(0++)++$" in the obvious way.)
The strategy to compute $m+n$ is: apply the second axiom $m$-many times to get $(0+n)++...++$, and then apply the first axiom once. It's a good exercise to show by induction that this always works. (Note that this proof will take place in the metatheory, since it relies already on a notion of "natural numbers".)
Note that a more common notation for "$x++$" is "$S(x)$"; I've used the former (since that's what the OP uses), but I strongly prefer the latter since it allows us to write things like "$SSSSSSSS(0)+SSSS(0)$" unambiguously and (relatively) shortly.