Proof Check Lemma 2.2.10 in Tao

You have proved the uniqueness. Also you need to show the existence of such $b$.

To do that, you need to consider the statement $$P(a)\equiv \text{ there exists a } b \text{ such that } b\!+\!\!+=a \text{ whenever } a\ne0$$ because $a$ is positive. Then induct on $a$.

Note that in certain step, the statement is vacuously true.

We have to prove $$a\ne0\implies\exists b\in\mathbf N,\;\;b\!+\!\!+=a.$$ So, we induct on $a$. The base case $a=0$ is vacuously true. Now suppose inductively that the claim is true for $a$; we need to show the claim for $a\!+\!\!+$, i.e., $b'\!+\!\!+=a\!+\!\!+$ for some natural number $b'$. Thus, by induction hypothesis, we have $b\!+\!\!+=a$. Applying the increment (by Sustitution axiom of equality) we obtain $(b\!+\!\!+)\!+\!\!+=a\!+\!\!+$. Defining $b':=b\!+\!\!+$ the claim follows.