Is that true though? As in, is it really that dangerous? It seems that you'll dissipate power equal to the inefficiency times the nominal charging power, so something like 5V x 2A x inefficiency (inefficiency being 1-efficiency), which will probably be of order a watt.
I can use my car battery to charge itself without any issues --- I just plug the red terminal to itself, and same with the black, which is to say, a battery is always connected in a way that "charges itself."
I think the key is that the battery probably isn't really playing a big role in OOP's setup --- electricity doesn't "go through the battery," it just goes from the charging input to the power output circuits, with the additional power (due to inefficiency) being provided by the battery.