For any locally free sheaf $\mathcal{E}$ on $X$, the projective bundle $\mathbb{P}(\mathcal{E}\oplus \mathcal{O}_X)$ possesses an open subset $\mathbb{V}(\mathcal{E})$ whose closed complement is isomorphic to $\mathbb{P}(\mathcal{E})$ and corresponds to the divisor associated to the sheaf $\mathcal{O}_X(1)$.

In your case $\mathrm{dim}(\mathbb{P}(\mathcal{L}\oplus \mathcal{O}_X))= 1$, so $\mathbb{P}(\mathcal{L})$ is the divisor at infinity (a point) and $\mathbb{V}(\mathcal{L})$ is the complementary line, see Fulton-Lang "Riemann-Roch Algebra" IV, 1. Moreover, if you consider the canonical map $\pi \colon \mathbb{P}(\mathcal{L}\oplus \mathcal{O}_X)) \to X$ then $\pi^*\mathcal{L}$ will correspond to a divisor of degree one in your projective line, again isomorphic to the divisor associated to the sheaf $\mathcal{O}_X(1)$.