If you don't know the type of DC/DC converter in supply mode, it is always a good idea to add a diode as shown in the picture.
The benefit is that you will protect your DC supplies from drawing current from another supply.
The drawback is that because there is a voltage drop on your diode (~0.7 V typical), there will be power loss of:
W = 0.7 * I;
So say, your lamp draws 300 mA, then your USB power loses 210 mW on that USB diode (it will get hot - too hot if you don't use a heat sink!).
However, there is a bigger problem in this circuit. If you disconnect from USB, your power bank will be discharging because it is lighting the lamp, right? But, since it is discharging, battery voltage will drop and this will trigger the battery charger to start charging. So, your powerbank will be discharging to charge itself! A cheap powerbank usually has 75% efficiency with their supply/charger units so a "discharge to self charge" situation has:
1 - 0.75 * 0.75 = ~45% Power loss!!
Your powerbank will not be as productive as it is supposed to be. So, this is not a good idea.
However, you can try this circuit:
As you can see, this way, your USB port and powerbank output are protected. Your powerbank is only charged when there is power at USB port. Also it is charged without a series diode on the way, which helps with efficiency.
Also hint: try to pick USB diode with less on resistance a.k.a. ~0.6V while let the powerbank be higher at ~0.7V. This is because you want your USB to both charge the powerbank and light the lamp. If you don't do this, then it is possible that powerbank lights the lamp, and is charged from input by USB. This is inefficient!