The battery has two ends: a positive end and a negative end. The positive end is at a higher potential than the lower end. The current can flow only when the charge carrier leaving the positive end can return back to the negative end (conventional current).
Note: * It's not the charge carrier that should return actually. For more info, read this post in this site.. I don't know whether you can understand what electric field waves are or not, so I didn't mention the term here. Once you have understood what electric fields are, you should say its "electric field waves that should return: not the charge carriers".*
With this in mind, let’s come to your question. When the switch is opened, there is no way for the current to return back to the battery. So the light doesn't glow.
But once the switch is closed, there is a complete path for the current to return back to the battery. The current is flowing through the bulb because there is a potential difference between the ends of the bulb.
simulate this circuit – Schematic created using CircuitLab
I have modeled the bulb to be a simple resistor (ignore the resistance). The -ve end of the battery is always at 0V potential. So the resistor has a potential difference of 1.5V across it, when the switch is closed.
To clarify what the switch actually does, consider the bulb is removed. So, there is no way for a charge carrier to return back. So just closing the switch doesn't allow the current flow. What actually happens is, the switch connects two of its ends, so both of its ends are at the same potential. The flow of current is determined by the condition I have mentioned above. How much current flows? Well, that depends on the total resistance offered by the whole circuit. For ideal cases, you can ignore the wire's resistance, battery's internal resistance. But for the practical case, they must be taken into account.