You are looking at two different bits of information. Both of which can be true.
Minimum supply voltage is 8.7V. This is the lowest the input voltage can be where the performance of the IC is guaranteed.
Meanwhile UVLO is in the range 8.45-8.9V. Meaning that somewhere between 8.45 and 8.9V the IC will turn off due to low input voltage.
If the input voltage is an ideal 8.71V (zero ripple, no drops whatsoever), the IC is guaranteed to work. However, if you happen to have an IC which due to the variations always present in manufacture, has a UVLO of 8.72V, the IC will go into its undervoltage lock out (doing whatever it does there). This is correct. The IC is functioning as it should. On the other hand, if you happen to have an IC which due to the variations always present in manufacture, has a UVLO or 8.69V, the IC will operate as expected.
Li-ion cells operate on a range of 3-4.2V (roughly, different cells vary slightly). A pair of them in series (as the IC you’re talking about is designed for) will therefore be 6-8.4V, so to charge them, you need a higher than 8.4V. In my experience chargers tend to be 0.3-0.5V more than the fully charged battery pack, to make sure you always have some push to get the last bit into the cell. So operating down to 8.7V makes sense for the IC, any lower than that, and you’re not charging the cells properly. Which means the manufacturer probably hasn’t bother checking lower than that.
Going back to your issue: you’re trying to reduce the power dissipation of the charge IC. It’s an IC, it is rated to use around 0.75 mA to 4 mA, this is pretty low compared to the power the li-Ion battery it is charging, why are you wanting to reduce this? If you’re really desperate to reduce power draw, your best bet is to search for a new IC.