Show that every morphism $f\colon \mathbb{A}^1\setminus\{0\}\to \mathbb{P}^1$ can be extended to a morphism $\mathbb{A}^1\to \mathbb{P}^1$
This is Exercise 5.7(a) of Gathmann's 2021 notes of Algebraic Geometry. Morphism means morphism of prevarieties.
The author gave the definition of $\mathbb{P}^1$ by gluing two affine lines $X_1= \mathbb{A}^1$ and $X_2=\mathbb{A}^1$ via $x\mapsto \frac{1}{x} $ (Example 5.5(a)) and then gave this exercise right after this definition. So it would be the best if there is an argument using only the definition...
Writing the points in $\mathbb{P}^1$ using the usual convention $[x_0:x_1]$, I can see that a morphism $f\colon\mathbb{A}^1\setminus\{0\}\to \mathbb{P}^1$ should be of the form
$$ f(x) = \begin{cases} [g(x):1] & x\in f^{-1}(X_1)\\ [1:h(x)] &x\in f^{-1}(X_2) \end{cases}$$
for some $g\in \mathscr{O}_{\mathbb{A}^1\setminus \{0\}}(f^{-1}(X_1))$ and $h\in\mathscr{O}_{\mathbb{A}^1\setminus \{0\}}(f^{-1}(X_2)) $.
I know (guess) that some argument about the degree of the both sides of the equation
$$ g(x)h(x)=1,\quad\forall x\in f^{-1}(X_1\cap X_2), $$
would tell that $g(x)=a_nx^n$ and $h(x)=a_n^{-1}x^{-n}$ or something and then we can use these to define the desired extension $\mathbb{A}^1\to \mathbb{P}^1$. However, elements in $\mathscr{O}_{\mathbb{A}^1\setminus \{0\}}(f^{-1}(X_i))$ could be wild, and I have no clue how to deal with this.
If instead of $\mathscr{O}_{\mathbb{A}^1\setminus \{0\}}(f^{-1}(X_i))$, we have $g,h\in \mathscr{O}_{\mathbb{A}^1\setminus \{0\}}(\mathbb{A}^1\setminus \{0\})=K[x]_x$, then things become easy: we know that $g$ and $h$ are polynomials on $x$ and $\frac{1}{x}$ and then the argument about degree applies.
I guess that maybe some property of $f$ will resolve this, or maybe replacing $f^{-1}(X_i)$ wisely to some other sets will resolve this, but I don't see how.
Thanks in advance for any help.