This is an answer to Georges' updated question at the end of his post.
An equivalent formulation of the question is the following:
Question Let $L$ be an ample Cartier divisor on a projective scheme $X$ and suppose there exist effective divisors $D_1, D_2$ such that $L\sim D_1-D_2$. Then is it true that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is affine?
I think this is true in some cases, but not in general.
Claim 1 The answer to the question is YES if $X$ is a projective curve.
Proof Both $D_1$ and $D_2$ are effective and hence ample and similarly so is $A=D_1+D_2$. Clearly $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)=X\setminus {\rm supp}\, A$, which is affine. $\square$
Claim 2 There are many examples for smooth projective varieties for which there exists $L, D_1, D_2$ as above such that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is not affine. In fact, this happens on any smooth projective surface containing a $(-1)$-curve.
Remark I am pretty sure one does not need smoothness and there are also singular examples. (Actually the example below only needs one smooth point.)
Proof Let $Y$ be an arbitrary projective variety (reduced) of dimension at least $2$ and $H$ an effective (very) ample Cartier divisor on $Y$. Let $\sigma : X\to Y$ be the blow up of a smooth point $p\in Y$ that is not contained in $H$ and let the exceptional divisor of $\sigma$ be $E\subset X$.
Then for some $m>0$ positive integer, $L=m\sigma^*H-E$ is ample. (I suspect that most people know this, but if you need a hint for this statement, an explicit estimate on $m$ can be found in Lemma 2 of this answer to another MO question.)
Now let $D_1=m\sigma^*H$ and $D_2=E$. Notice that by the choice of the point that was blown up, $D_1$ and $D_2$ are disjoint. It follows that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)\simeq (Y\setminus {\rm supp}\, H)\setminus \{p\}$. Furthermore, since $H$ is ample on $Y$, it follows that $Y\setminus {\rm supp}\, H$ is affine, and hence $(Y\setminus {\rm supp}\, H)\setminus \{p\}$ is not. $\square$
It is actually true, that for any line bundle there always exists a rational section for which the complement of its divisor is affine.
Claim 3 Let $L$ be an arbitrary Cartier divisor on a projective scheme $X$. Then there exist effective very ample divisors $D_1, D_2$ such that $L\sim D_1-D_2$.
Proof Choose an arbitrary ample Cartier divisor $A$ on $X$. For large enough $r_1\gg 0$ $L+r_1A$ is basepoint-free by the definition (or one of the basic properties depending on what you choose as definition) of ampleness. Then for an even larger $r\gg r_1$ we may assume that $L+rA$ is both basepoint-free and ample and hence very ample and also that $rA$ is very ample as well. Now choose $D_1=L+rA$ and $D_2=rA$. $\square$
And we get as an easy consequence:
Corollary With the notation of Claim 3, we may choose $D_1$ and $D_2$ such that $X\setminus \left({\rm supp}\,D_1 \cup {\rm supp}\,D_2\right)$ is affine.
Proof Replace $D_1$ and $D_2$ with general members of their complete linear systems. Then we may assume that they do not have a common component and hence ${\rm supp}\,(D_1+D_2)={\rm supp}\,D_1 \cup {\rm supp}\,D_2$. Since $D_1+D_2$ is also ample, this proves that claim. $\square$
