Proving existence and uniqueness of a function $f':\mathbb{N}\rightarrow A$ Suppose $A$ is a group with operation $\times$.
Let $f:\mathbb{N}\rightarrow A$ be a function where, for any natural number $m,n$
$f(m+n)=f(m)\times f(n)$
We want to show that this can be extended to a function $f':\mathbb{Z}\rightarrow A$ where, for any integer $m,n$
$f'(m+n)=f'(m)\times f'(n)$
Show that such a $f'$ exists and that it is unique.
So far, I have $f'(n)=f(n)$ when $n>0$ and $f'(n)=1$ (identity) when $n=0$ and $f'(n)=f(-n)^{-1}$ for $n<0$
Does this function work, if not, whats missing?
How do I demonstrate that the function i defined works and that its unique?
 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 $g=f'$.
