The $1-2\alpha$ is not because you calculate the CI for each group separately. It is because you calculate the "inequivalence" to the upper and to the lower end separately. The parameter $\theta$ lies in the equivalence interval $[\epsilon_L, \epsilon_U]$ iff $$\theta \geq \epsilon_L \wedge \theta \leq \epsilon_U.$$
Each part is tested separately by a one sided test at level $1-\alpha$. Only if both tests are significant, we can conclude equivalence. (This is the very intuitive intersection-union-principle.) Turning this into a single confidence interval, we must remove $\alpha$ from both the upper and the lower probability mass of the CI. So we end up with $1-2\alpha$. The TOST-CI is simply the intersection of the one-sided CIs.
By the way, it is still possible to do the TOST with a $1-\alpha$ CI, but it would be unnecessarily conservative.