Assuming that $\otimes$ preserves finite colimits in each separate argument, I think it's true in general.
Since it's very easy to get twisted around here, let me abstract the situation a bit. Recall that if $\alpha: F' \to F$ is a map between left adjoints in a bicategory (with right adjoints $G'$ and $G$ respectively), then the mate $\alpha^\ast: G \to G'$ is defined by the composite
$$G \stackrel{\eta' G}{\to} (G'F')G \stackrel{(G'\alpha) G}{\to} (G'F)G \cong G'(FG) \stackrel{G'\varepsilon}{\to} G'$$
where $\eta'$ and $\varepsilon$ are the appropriate unit and counit of adjunctions. Here of course we are thinking of our monoidal category as a one-object bicategory.
If $\pi: \mathcal{O} \to L$ is an epi, then I claim its mate $\pi^\ast: L^{-1} \to \mathcal{O}$ is a mono. For, if $f, g: A \to L^{-1}$ are distinct arrows, then the two composites
$$A \stackrel{\overset{f}{\to}}{\underset{g}{\to}} L^{-1} \stackrel{\pi^\ast}{\to} \mathcal{O}$$
are equal to the composites
$$A \cong \mathcal{O} \otimes A \stackrel{\pi \otimes A}{\to} L \otimes A \stackrel{\overset{L \otimes f}{\to}}{\underset{L \otimes g}{\to}} L \otimes L^{-1} \stackrel{\varepsilon}{\cong} \mathcal{O}$$
by the usual yoga of mates and adjunctions; note that $L \otimes f$ and $L \otimes g$ are distinct since tensoring with $L$ is an equivalence, and note that $\pi \otimes A$ is an epi because tensoring with $A$ is right exact. So these two composites are distinct. Hence $\pi^\ast$ is monic.
On the other hand, the mate $\pi^\ast$ is also epic, because it is given by the composite
$$L^{-1} \cong \mathcal{O} \otimes L^{-1} \stackrel{\pi \otimes L^{-1}}{\to} L \otimes L^{-1} \stackrel{\varepsilon}{\cong} \mathcal{O}$$
where the middle arrow is epi again because tensoring with an object is right exact.
Since $\pi^\ast$ is mono and epi in an abelian category, it is an isomorphism, and therefore its mate $\pi$ is also an isomorphism.
