If you are specifically interested in the case where $L=\mathcal{O}_{X}(C_{1}-C_{2}),$ all that is needed for your determinant bundle argument to work is the vanishing $H^{0}(\mathcal{O}_{X}(2C_{1}-2C_{2}))=0.$ But this follows from taking cohomology in the exact sequences $$0 \rightarrow \mathcal{O}_{X}(-2C_{2}) \rightarrow \mathcal{O}_{X}(C_{1}-2C_{2}) \rightarrow \mathcal{O}_{C_{1}}(C_{1}-2C_{2}) \rightarrow 0$$ $$0 \rightarrow \mathcal{O}_{X}(C_{1}-2C_{2}) \rightarrow \mathcal{O}_{X}(2C_{1}-2C_{2}) \rightarrow \mathcal{O}_{C_{1}}(2C_{1}-2C_{2}) \rightarrow 0$$
EDIT: Up to now we have only established that the elements of ${\rm Hom}(E,E \otimes L)$ have rank at most 1 on each fiber. If the restriction of $E$ to $C_{1}$ is $\mathcal{O}_{C_1}(a) \oplus \mathcal{O}_{C_1}(b)$ for $a,b \in \mathbb{Z}$ satisfying $|a-b| \leq 1,$ then the simplicity of $E$ and the sequence $$0 \rightarrow E \otimes E^{\ast}(-C_2) \rightarrow E \otimes E^{\ast} \otimes L \rightarrow E \otimes E^{\ast} \otimes \mathcal{O}_{C_1}(-2) \rightarrow 0$$ give you what you want; I don't see right away how to proceed without extra assumptions on $E.$
If you are specifically interested in the case where $L=\mathcal{O}_{X}(C_{1}-C_{2}),$ all that is needed for your determinant bundle argument to work is the vanishing $H^{0}(\mathcal{O}_{X}(2C_{1}-2C_{2}))=0.$ But this follows from taking cohomology in the exact sequences $$0 \rightarrow \mathcal{O}_{X}(-2C_{2}) \rightarrow \mathcal{O}_{X}(C_{1}-2C_{2}) \rightarrow \mathcal{O}_{C_{1}}(C_{1}-2C_{2}) \rightarrow 0$$ $$0 \rightarrow \mathcal{O}_{X}(C_{1}-2C_{2}) \rightarrow \mathcal{O}_{X}(2C_{1}-2C_{2}) \rightarrow \mathcal{O}_{C_{1}}(2C_{1}-2C_{2}) \rightarrow 0$$