There are few mix-ups in this question.
First,
"two data pins seem to be connected to an integrated circuit (I
initially thought they would be shorted)"
The shorted D+/D- data pins is a signature of CHARGER/power provider, not consumer.
so is it possible that the integrated circuit is part of the battery
management system (BMS)?
The BMS is not a correct term for the old tablet, it is usually a function built-in the internal battery itself.
What would happen if I connected only the positive and negative pins
and connected the data pins to pull-down resistors and then to ground?
Nothing will happen, no charging. You need to connect CC pins, not "data pins" to ground with 5.1k to get VBUS out of Type-C link. Then the tablet will charge, but likely just at 500mA level.
The D+/D- wires in your tablet are likely used to either understand the charging signature of the original charger (it could be BC2.1 old USB specifications) and take more current than a standard USB2 port is obligated to supply (500mA), or used to inform the external charger about enhanced charging capability of the tablet (like Quick Charge protocol).
In any case the replacement of u-USB with Type-C receptacle will not be able to engage any enhancement in charging of the tablet. The Rd on both CC pins will only enable a Type-C charging source to enable VBUS, but the amount of current that the tablet is going to consume depends on CC voltage levels, which is not built into the old tablet. This is because that the amount of consumption is determined by the tablet.
What you can try is to attach the Type-C receptacle with both 5.1k pull-downs, and internally short D+/D-. This will likely mimic the "Chinese charger signature" for the tablet, and it likely will be able to charge its battery at 1.5A level. It is very likely that any Type-C port will be able to supply 1.5A, since it is a requirement today.
