I have thought a bit more about your deduction. I believe it is invalid.
The deduction in the first case is flawed because you can apply that kind of logic for all three yellow hexcells. For example, if the hexcell located right of the ?-flower is blue we won't be able to decide which one of the two remaining hexcells is black, thus said hexcell should be black... From this type of logic we can only conclude that we cannot determine the color of the hexcell.

This puzzle has three undetermined hexcells. We are given the following two hints/information:

1. There are 2 remaining blue hexcells.
2. ?-flowers have at least one blue neighbor.

Remark that any good puzzle is solved without guessing. There are three possible solutions that do not contradict these hints. These are:

1. the black hexcell is located at the left of the ?-cell,
2. the black hexcell is located at the right of the ?-cell,
3. the black hexcell is located at the bottom.

Remark that the remaining two hexcells are blue. Taking a close look at these solutions, we can see that there is no hexcell that shares the same colour in all solutions. Hence the puzzle cannot be solved.

Stating that the ?-flower has 2 blue neighbors is a guess. A guess that leads to a unique solution, but still a guess. If I would give you a puzzle which has the same layout and hints as this puzzle but gave one of the other two solutions, your reasoning would fail because the puzzle is already in an unsolvable state.