 Solution

The assertion is false. The Picard group of the Hirzebruch surface is freely generated by the negative section $S$ and a fiber $F=D_1$. Pullbacks from ${\mathbb{P}}^1$ form only the subgroup $\mathbb{Z}F$. For example, $\mathcal{O}(S)$ cannot be a pullback: its restriction to a fiber has degree $S\cdot F=1$, whereas every pullback from the base restricts trivially to every fiber.