Clearly $0$ and $<$ are definable in this model. By way of contradiction assume * is definable. I look at the model extension $\mathbb{N}*\mathbb{Z} - \{0\}*\mathbb{Z}_-$ with $(a,b)+(c,d)=(a+c,b+d)$, denote $n=(0,n)$ and $x=(1,0)$. So this is the closure of $\mathbb{N}$ with some x that satisfies $\{x>n| n \in \mathbb{N}\}$.
I claim that $x^2=x*x$ is not defined there: if $x^2=ax+b$ then $b=0$ (otherwise it would be divisible by $x$ but $x>b$) and then reduce it to $a=x$, a contradiction. However, the closeness of the multiplication (just like associativity, divisibility rules, etc.) is a first-order-logic definable.
Edit:
The result follows from the deep differences between Presburger Arithmetic and $(\mathbb{N},+,*)$. My proof below is wrong as shown in Alex Kruckman's answer but something similar might work.
No, your proof is not valid.
Let's denote your model by $\mathcal{N}$. If I understand your notation, it looks like $\mathcal{N}$ consists of a copy of $\mathbb{N}$, followed by a sequence of copies of $\mathbb{Z}$. And $x$ is the "$0$ element" of the first copy of $\mathbb{Z}$. Right?
Now the biggest problem with your argument (you also have some serious imprecision in the second paragraph) is that in order to conclude anything about definability in $\mathbb{N}$ from looking at $\mathcal{N}$, you need $\mathcal{N}$ to be elementarily equivalent to $\mathbb{N}$ - and it's not.
To see this, note for example that $\mathbb{N}$ satisfies the sentence $$\forall x\, \exists y\, (y+y = x\lor y+y= x+1).$$ (You can rewrite the above sentence to remove the explicit reference to $1$, since $1$ is a definable element.) This says that every element or its successor is divisible by $2$. In your model $\mathcal{N}$, neither $x$ nor $x+1$ is divisible by $2$.
A variant on this argument shows that any model elementarily equivalent to $(\mathbb{N},+)$ must have order-type $\mathbb{N} + L\times \mathbb{Z}$, where $L$ is a dense linear order without endpoints. This suggests the question of whether $\mathbb{N}+\mathbb{Q}_+\times \mathbb{Z}$ (with its natural additive structure) is elementarily equivalent to $(\mathbb{N},+)$. I don't know the answer off the top of my head, and it would take some work to establish this. But if it is, it's possible an argument like the one you suggest could work using this model.
Of course, one can alternatively prove non-definability of multiplication by appealing to bigger theorems. The theory of $(\mathbb{N},+)$ is (a reduct of) Presburger Arithmetic, which is well-known to be decidable. But the structure $(\mathbb{N},+,*)$ is undecidable, by Gödel's theorem.