Electrical – Powering 12V 2A device from high-power USB portable charger

That charger takes any voltage between about 2V and less than 5V and boosts it to 5V.

You could use that circuit with a 2V to say 4.5V solar panel BUT you can instead just use a solar panel directly.

A solar panel with full sun loaded voltage of 6V or more will charge a USB cellphone either directly or with the addition of a series diode. (Diode needed ONLY if cellphone battery voltage appears on charging input. )

For 'safety' you should also connect a 5V1 zener-diode (5.1 Volt) across the solar panel output so that Vout can never be > 5.1V.
Zener Wattage = Wmp panel = max panel power in full sun.

Charge rate is related to panel Wattage.
Panel peak watts in full sun = Wmp.
The panel wattage for maximum charge needs be no more than :

Vpanel_loaded_at_full_sun / Ah_of_cellphone_battery

eg a panel with Vmp=6V and charging a 1000 mAh = 1 Ah cellphone battery need have a Wattage of no more than 6V/1Ah = 6 Watts. This would charge the battery at maximum possible rate. Higher wattage panels will heat the zener and lower wattage panels take longer to charge.

What can I do to make the charger output power on low loads?

You probably can't do anything reasonable. Charging power banks are designed to cut off power when they think the charge is done, when load drops below certain cut-off level. This condition will happen every time you connect-disconnect your project while debugging it.

1. you can load it with proper resistor, but it will drain the bank battery quickly.

2. you can reverse engineer your powerbank, determine what kind of PMIC (power management IC) is used, and disable this cut-off function if possible, but it is very doubtful.

3. You need to design your own PMIC on the top of existing circuit, replace it with power management you need. This new design must have all features to charge and protect the powerbank battery, but will have no cut-off function.