Let $k$ be an algebraically closed field and $\mathbb{A}^1_k:=\operatorname{Spec}(k[x])$ be the affine 1-space. Why is then $\mathbb{A}^1_k \setminus \{0\} \simeq \operatorname{Spec}(k[x,x^{-1}])$ ?
My ideas: Let $S:= \{ x^n | \: n \in \mathbb{N}_0 \}$. Then $S$ is multiplicatively closed and $S^{-1}k[x] = k[x,x^{-1}]$. The prime ideals of the localisation of $k[x]$ at $S$ are exactly the prime ideals of $k[x]$ which don't intersect $S$. So $(0)$ and $(x-a)$ for $a \neq 0$. Is this what is meant by $\mathbb{A}^1_k \setminus \{0\}$ ?
Really, this is a matter of definitions. If we regard $\Bbb{A}^1_k$ as the set $k$ with its Zariski topology, then $\Bbb{A}^1_k\setminus \{0\}$ is an open subset $U$ of $\Bbb{A}^1_k$ and the regular functions on $U$ are by definition quotients $f(x)/g(x)$ such that $f,g\in k[x]$ and $g(p)\ne 0$ for all $p\in U$. In particular, the only admissible $g(x)$ are of the form $ax^n$ for $a\in k$ and $n\in \Bbb{N}$. So, $\mathcal{O}(U) = k[x,x^{-1}] = k[x]_x$.
The scheme $\Bbb{A}^1_k$ is defined to be $\mathrm{Spec}\:k[x]$. Then by $\mathbb{A}^1_k\setminus \{0\}$ you might mean $\mathrm{Spec}\:k[x]\setminus \{(x)\}$, where $(x)$ is the ideal of functions corresponding to the closed point $0$ in the classical picture. In that case, note that $\{(x)\}$ is closed and so $\mathrm{Spec}\:k[x]\setminus \{(x)\}$ is open, defined to be the complement of $V(x) = \{(x)\}$. In particular, by the definition of the sheaf of functions on an affine scheme, the functions are $k[x]_{x} = k[x,x^{-1}]$.
In short, after suitable interpretation the answers are the same. You just need to be careful what language you mean.
Note that the classical variety $\mathbb{A}^1_k\setminus \{0\}$ in the first paragraph is not the same space as $\mathrm{Spec}\:k[x,x^{-1}]$ because the latter space has a generic point $\xi$ corresponding to the zero ideal which is dense in the whole space. That point is not closed, so the spaces are not even homeomorphic. The latter is the soberification of the former.