Solution:

We directly exhibit a sequence of moves that leads to having all of the first 2019 light bulbs on.

- If the light bulb numbered 2019 is off, we choose $a=2019$ and $b=673$, so that $\frac{a}{b}=3$ is a prime number, and we change the state of both light bulbs. Now light bulb 2019 is on.

- If light bulb number 1 is off, we choose $a=2$ and $b=1$ and change the state of both. Now light bulb number 1 is on.

- If light bulb number 2 is off, we choose $a=4$ and $b=2$ and change the state of both. Now light bulb number 2 is on.

- If light bulb number $k$ is off, with $1 \leq k \leq 2018$, we choose $a=2k$ and $b=k$ and change the state of both. Now light bulb number $k$ is on.

- If light bulb number 2018 is off, we choose $a=4036$ and $b=2018$ and change the state of both. Now light bulb number 2018 is on.

Note that every light bulb that is turned on during a move is then never changed in state again by subsequent moves: indeed, light bulb 2019 is changed in state only on the first turn, because in all subsequent turns light bulbs of the form $k$ or $2k$ with $1 \leq k \leq 2018$ are used and 2019 is odd, while if a light bulb $k$ with $1 \leq k \leq 2018$ has its state changed, in subsequent turns only light bulbs with numbers greater than $k$ will be considered.

Note: this solution works only if the moves are applied in the order described above, otherwise one risks turning off again a light bulb that had previously been turned on.