The part is (allegedly) tested for 1500V RMS AC voltage (one minute), which is 2.1kV peak but there's no indication of the rating for continuous operation. That's not really a very impressive test voltage for 1050V DC continuous operation. You should probably take precautions to limit the damage from a failure. More likely they actually test it for a much shorter time at a higher voltage so the probability of it passing a 1 minute test at 1500VAC RMS is very high.
The similar Vishay (nee Infineon, nee Siemens) IL300 datasheet is here. It is tested for 1s at 5300 VRMS (about 7500V peak).
You can see some typical application circuits in the above-linked datasheet that will help answer your question about working from the shunt. You need op-amps on both sides, and you need power supplies on both sides.
The (transfer) accuracy of the part you chose is much better than the IL300, but even so you might want consider a more expensive isolation amplifier that has some of that stuff built-in and perhaps uses a different isolation method that is explicitly rated for the service you have in mind.
If you really want to guarantee reliability and cannot find an isolator with an appropriate rating and guarantee from the manufacturer, you can use a fiber optic cable to isolate the signal (digitize it first). Provided the materials used are appropriate, almost any reasonable isolation voltage should be possible.
Or, lose the idea of the shunt entirely and use a DC closed-loop Hall effect pickup with appropriately insulated wire. That would be my best suggestion!