Hello,
From what i can gather on the data sheet, with your circuit if the input goes to -12v that will put about -0.6v at pin 3, so that's 11.4v across 15k which is 760ua which is not destructive to the IC chip itself, however, the output during this negative excursion may go to the highest voltage level possible which is equal to +Vcc. If your control logic can handle this condition you're ok without changing anything at all.
A possible solution to the lack of a feedback resistor (and hence any hysteresis) is to put the feedback resistor on another board and bring out connections from pin 3 and pin 1 to that other board. #22 gauge buss wire is good for fixes like this, but i've also used #24 with success for years. Anything of a lighter gauge may snap if the board is moved around.
Another possible solution is to use an 1/8 watt resistor with leads covered with teflon sleeving. for maximum physical stability you could drill holes near the nodes to be connected and put the leads through, then run the leads (very short) under the board to the nodes and solder. The teflon sleeving method is used in industry a lot and the hole drilling method is even more stable if of course you have the room on the board to drill holes right next to the pins to be connected to.