Let $G$ be an abelian group and $x\in G$. Then there exists a unique homomorphism $f:\mathbb{N}\rightarrow G$ with $f(1)=x$ and $f(n)=n\cdot x$ for all $n\in\mathbb{N}$.
Since $x$ is invertible, there also exists a unique homomorphism $g:\mathbb{Z}\rightarrow G$ such that $g(1)=x$ and $g$ coincides with $f$ on $\mathbb{N}$.
For arbitrary $n\in\mathbb{Z}$, what is the $g(n)$ equal to? If $n$ is a negative integer, then I'm afraid the expression $n\cdot x$ is nonsense: it ordinarily denotes the composition of an ordered sequence of $n$ terms all whose terms are equal to the same element $x\in G$. But if $n=-m$ for some $m\in\mathbb{N}$, what could possibly be meant by "a sequence of $-m$ terms whose terms are equal to the same element $x\in G$"?
For example, let $G:=\mathbb{Z}$ and $m\in\mathbb{Z}$. Then there exists a unique homomorphism $g:\mathbb{Z}\rightarrow\mathbb{Z}$ such that $g(1)=m$ and $g$ coincides with $f$ on $\mathbb{N}$. Let $n$ be an arbitrary element of $\mathbb{Z}$, what is $g(n)$ equal to? Is it $n\cdot m$? But what if $n$ is negative?
"$-$" in an abelian group usually denotes the inverse. Thus $-nx$ for $n$ positive is equal to the additive inverse of $nx$. For $n$ a negative integer, $$nx = -((-n)x)$$ Note that $$-(nx) = n(-x)$$ so it's a sequence of $n$ terms of $-x$.