Let $\mathsf{ZFC}'$ be the extension of $\mathsf{ZFC}$ containing the constant symbol $\Bbb N$, which we take to represent the natural numbers. In order to say that $\mathsf{ZFC}'$ is a definitional extension of $\mathsf{ZFC}$ we need to find a formula $\phi$ in the language of $\mathsf{ZFC}$ with a single free variable $\upsilon$ such that $\mathsf{ZFC}\vdash\exists!\upsilon\phi$ and, for any formula $\psi$ containing $\Bbb N$, $\mathsf{ZFC}'\vdash\psi$ iff $\mathsf{ZFC}\vdash\exists!\upsilon(\phi\land\psi(\Bbb N\mapsto\upsilon))$ or equivalently $\mathsf{ZFC}\vdash \forall\upsilon(\phi\implies\psi(\Bbb N\mapsto\upsilon))$.
However, since there can be no recursive axiomatization of $\text{Th}(\Bbb N)$ – provided that $\Bbb N$ is truly the set of natural numbers – there can be no formula $\phi$ uniquely characterizing $\Bbb N$ (up to isomorphism). So either $\Bbb N$ is not the set of natural numbers, $\mathsf{ZFC}'$ is not an extension by definitions, or $\mathsf{ZFC}$ is inconsistent.
Best Answer
You write:
This is incorrect. Very broadly speaking, what we can say is that $\mathsf{ZFC}$ (being recursively axiomatizable) must not be able to settle all questions about $\varphi$. But this has nothing to do with $\mathsf{ZFC}$ proving that exactly one thing satisfying $\varphi$ exists or that thing corresponding appropriately to $\mathbb{N}$. For example, $\mathsf{ZFC}$ also proves "There is exactly one set $x$ which is $\emptyset$ iff $\mathsf{CH}$ holds and is $\{\emptyset\}$ iff $\mathsf{CH}$ fails," while not settling the question of whether this unique object is empty.
There are various senses in which $\mathbb{N}$ is "hard to pin down" and various other senses in which $\mathbb{N}$ is "easy to pin down;" you have to be very careful about which sense is being used when applying a given theorem.