So I try to build a deterministic pushdown automata. I am not sure whether this is automata is indeed deterministic though. In this case, $e$ denotes the empty word. I need to seperate the amount of $a$'s I choose, but since the automata needs to be deterministic, I can't choose the same transition from $q_1$ to $q_2$ and from $q_1$ to $q_3$. Is it valid to use the empty word here? Or is it still non-determionistic since the empty word basically doesn't count?
You can't have empty transitions in a deterministic automaton.
"Deterministic" means that there must only be one transition that is even possible to follow in every imaginable configuration of the machine.
In your case it is easy to see that in $q_1$ with the next symbol being $a$ and nothing on the stack (it looks like you're not using the stack at all?) both the ransition to $q_2$ and the one to $q_3$ could be taken. So it's not deterministic.
Usually, to enforce this requirement_ one doesn't allow empty transitions in a deterministic automaton. Sometimes, for PDA's, one allows transitions without symbols if the transition at least pops an element off the stack, but which transition to take still need to be determined locally by looking jointly at the next input symbol and the one at the top of the stack.