Is algebraic dependency decidable?
A model of computation allowing real numbers is the Real-RAM model. In this model, it is impossible to even decide if a number is rational. Indeed, the time-T halting sets are semialgebraic. If any one of them has a non-empty interior, then the time-T yes-set and time-T no-sets are semialgebraic subsets with no-empty interior, which is a contradiction. If none of the time-T halting sets is semialgebraic, then their union cannot be $\mathbb{R}$.

Answer to the previous version of the question:

If $a_1,\dotsc,a_k$ are algebraic numbers that are linearly independent over $\mathbb{Q}$, then $e^{a_1},\dotsc,e^{a_k}$ are algebraically independent over $\mathbb{Q}$. This is the Lindemann–Weierstrass theorem. So, if $a$'s are algebraic (as in the current version of the question), then the answer to your question is "yes".