In this case, $R^1\pi_*\mathcal{O}_T$ is still an invertible sheaf, but that is not necessarily true for other families having a double fiber. You can compute $R^\bullet\pi_*\mathcal{O}$ after a finite flat base change, e.g., by the squaring map $\Delta'\to \Delta$, $w\mapsto z = w^2$. The fiber product $T' = \Delta'\times_{\Delta} T$ is a non-normal surface. The normalization $\nu:\widetilde{T}'\to T'$ is an isomorphism away from the (reduced) double fiber $F$. Over $F$, the map is $\tilde{F}\to F$, an etale, degree $2$ morphism of elliptic curves. Thus there is a short exact sequence $$0\to \mathcal{O}_{T'} \to \nu_*\mathcal{O}_{\widetilde{T}'} \to (\nu_*\mathcal{O}_{\widetilde{F}}/\mathcal{O}_F) \to 0.$$ Since both $h^0$ and $h^1$ are zero on $\nu_*\mathcal{O}_{\widetilde{F}}/\mathcal{O}_F$, it follows that the induced map $R^\bullet \pi'_*\mathcal{O}_{T'} \to R^\bullet (\pi'\circ \nu)_*\mathcal{O}_{\widetilde{T}'}$ is a quasi-isomorphism. Thus, since $\widetilde{T}'\to \Delta'$ is a smooth, proper family of elliptic curves, it follows that $R^1\pi'_*\mathcal{O}_{T'}$ is an invertible sheaf. Thus also $R^1\pi_*\mathcal{O}_T$ is an invertible sheaf.