Proving existence and uniqueness of a function $f':\mathbb{N}\rightarrow A$

Your function works!

I'll omit the $\times$. And, as you used, $1 \in A$ will be the neutral element of the group $A$.

To see that it works we just need to do a case analysis.

Case $\boldsymbol{n>0}$ and $\boldsymbol{m>0}$: $$f'(n+m)=f(n+m)=f(n)f(m)=f'(n)f'(m).$$

Case $\boldsymbol{n=0}$ and $\boldsymbol{m\in \mathbb{Z}}$: $$f'(0+m)=f'(m)=f'(0)f'(m).$$

Case $\boldsymbol{n<0}$ and $\boldsymbol{m<0}$: $$f'(n+m)=f(-n-m)^{-1} = f(-m-n)^{-1} = (f(-m)f(-n))^{-1} =f(n)^{-1}f(m)^{-1}=f'(n)f'(m).$$

Case $\boldsymbol{n>0}$ and $\boldsymbol{m<0}$:

For that we will actually need to separate into 3 subcases. We will also use that $f'(n)=(f'(-n))^{-1}$.

Subcase $\boldsymbol{n+m>0}$: $$f'(n)=f'(n-m+m)=f'(n+m)f'(-m)=f(n+m)f(-m)=f'(n+m)f'(m)^{-1},$$ that is, $$f'(n+m)=f'(n)f'(m)$$

Subcase $\boldsymbol{n+m=0}$: $$f'(n) = f'(n+m-m) = f'(n+m)f'(-m) = f'(n+m) f(-m) = f'(n+m)f'(m)^{-1},$$ that is, $$f'(n+m)=f'(n)f'(m)$$

Subcase $\boldsymbol{n+m<0}$: see that $f'(n+m) = f'(-m-n)^{-1}$ and that $-n-m>0$, then we can use first subcase so that $$f'(n+m) = f'(-n-m)^{-1} = (f'(-m)f'(-n))^{-1} = f'(-n)^{-1}f'(-m)^{-1} = f'(n) f'(m).$$

And that covers every possible pair of integers $n,m$.

To see that it is unique, consider any function $g:\mathbb{Z} \to A$ such that $g(n)=f(n)$ for every $n>0$ and $$g(n+m)=g(n)g(m)$$ for every $n,m \in \mathbb{Z}.$

Then, putting $n=m=0$, we have $$g(0+0)=g(0)g(0) \implies g(0)=g(0)^2 \implies g(0)=1.$$

Moreover, let $n<0$ then using that $g(-n)=f(-n)$ and $g(0)=1$ we have $$g(n-n)=g(n)g(-n) \implies 1 = g(n)f(-n) \implies g(n)=f(-n)^{-1}.$$

And that is exactly how you defined $f'$. So $g=f'$.