Let $i: U=\mathbb A^2-0\to\mathbb A^2=X$ be the inclusion (over a base field $k$). It's well-known that $\mathbb A^2-0$ is not an affine variety and the restriction map $\mathcal O_X(X)\to\mathcal O_X(U)$ is an isomorphism. I want to know if
$\mathcal O_X\to i_*\mathcal O_U$ is an isomorphism?
To expand on what I said in the comment, here's a proof of the following statement.
Prop: Let $X$ be a scheme, and assume that $X$ is quasi-compact and quasi-separated i.e every intersection of two open affine subsets has a cover by a finite number of affine subsets. Let $f\in \Gamma(X, \mathcal{O}_X)$, and let $X_f=\{x\in X| f(x)\neq 0\}$, then $X_f$ is open in $X$ and $\Gamma(X_f, \mathcal{O}_X)\simeq \Gamma(X, \mathcal{O}_X)_f$.
Proof: To see $X_f$ is open, we may assume that $X$ itself is affine, but in that cas $X_f=D(f)$ which is open. Notice that $f$ is invertible in $\Gamma(X_f, \mathcal{O}_X)$ as it is everywhere locally. We thus have a well defined map $\Gamma(X, \mathcal{O}_X)_f\to \Gamma(X_f, \mathcal{O}_X)$, we're going to prove that it is an isomorphism.
When $X$ itself is affine, the result is obvious. Now take $X=\bigcup_i U_i$ with a finite numbre of $U_i$'s, all affine, we have a commutative diagram
$$\begin{matrix}0& \to & \Gamma(X, \mathcal{O}_X)_f&\to& \prod_i \Gamma(U_i, \mathcal{O}_X)_f&\to & \prod_{i,j}\Gamma(U_i\cap U_j, \mathcal{O}_X)_f\\ & & \downarrow& & \downarrow& & \downarrow\\ 0& \to & \Gamma(X_f, \mathcal{O}_X)&\to& \prod_i \Gamma(U_{i, f}, \mathcal{O}_X)&\to & \prod_{i,j}\Gamma({U_i\cap U_j}_f, \mathcal{O}_X)\end{matrix}$$ The arrow in the middle is an isomorphism since the $U_i$ are affine, therefore the arrow on the left is injective. If $X$ is separated you're done, as the arrow on the right also is an isomorphism.
In any case this means that if $Y$ is any scheme covered by finitely many open affine subsets $O_Y(Y)_g\to O_Y(Y_g)$ is injective.
Assume only the quasi separatedness of $X$, then the $U_i\cap U_j$ are themsleves covered by finitely many open subsets, and thus the arrow on the far right is also injective. But then an easy diagram chasing tells you that the arrow on the left is an isomorphism.
You can use the same ideas to prove that if $f: X\to Y$ is quasi-separated and quasi-compact then the direct image of a quasi-coherent sheaf will still be quasi coherent. I believe this is done in pretty much every reference.