This appears to be a duplicate of (half of) this question, but it received no attention so I'll try again.
Given a line bundle $L\to X$ on a scheme $X$ over a field $k$, I am to show that $L'=L\setminus z(X)$ the complement of the zero section is a torsor for the multiplicative group. Our definition of a $G$-torsor $T$ is, I think, somewhat special: one simply requires $T$ to be faithfully flat over $X$ and equipped with an action $G\times_X T\to T\times_X T$ which is an isomorphism. I'm working in functor-of-points style.
As I understand it, for $R$ a finitely generated $k$-algebra, $\mathbb{G}_m(R)=\coprod_{x\in X(R)} R^\times$. I can't quite get this to be the value of $L'(R)$. The most naive guess is that the fiber of $L'(R)$ over $x\in X(R)$ is simply the complement of zero in the projective $R$-module $x^*(R)$. But this is bad, and it seems the reason is that $L'$ has to be a subfunctor, so for instance elements of $L'(R)$ have to map to units under maps from $R$ to fields. I think this means elements of $L'(R)$ have to be generators, although I haven't written it out.
So it seems it would suffice to show that
The set of generators of a rank 1 projective $R$-module $P$ is a torsor for $R^\times$.
But if $a$ and $ra=r'a$ generate $P$, then all I seem to get from $r-r'$ annihilating $a$ is that $r-r'$ is nilpotent-since otherwise $P$ would be $0$-dimensional modulo some prime and not of constant rank $1$. What am I missing here? Did I run off the tracks entirely with the rank-1 projectives lemma?
OK, I got around the stumble in the last paragraph. If $a$ and $b$ generate some rank 1 projective $P$, then on some affine open cover of $\text{Spec}(R)$ (on which $P$ becomes a free rank-1 module) $a$ and $b$ are related by a unique coherent choice of units, so the sheaf condition on the group of units concludes the proof.