The schematic in the OP message is complete and clear, showing unipolar power supply and the opamp negative supply powered from the supply negative, same as the lower leg of the divider.
Complete and clear, you must be kidding. With no ground shown on the OP's schematic almost everyone would assume...
You know, adding the ground symbol somewhere does not change the circuit behavior the slightest. Ground is just one arbitrary circuit node, it has no special physical properties. SPICE simulators, though, AFAIK, require the ground symbol to be put
somewhere in order to simulate (which is understandable - SPICE simulators are able to report single node voltages (instead / in addition to voltage differences between two nodes), and for that, you need one specified reference point.
The schematic is complete because there are no floating nodes, opamp power is also connected directly (unipolarly) to the 12V voltage source. It's clear because the nets are properly labeled (inlucing opamp power rails), and drawn in a uncluttered way. I see absolutely nothing wrong* with it; I would use the ground symbol for the "V-" net, though, but it doesn't change anything.
*) wrong as in the way the schematic is drawn; functionally, it's clearly wrong because the circuit does not work, hence this topic!
You can add the missing ground symbol to any node (some are more
convenient than others, though). There is no need to "assume" anything about the schematic, it can be analyzed using the normal circuit analysis methods, like Kirchoff laws, opamp "rules of thumb", or simulated using SPICE, as long as the ground symbol is added to some node so SPICE can refer the voltages to that point.
I think your confusion is a good example why terminology such as "virtual ground" is dangerous. I'm baffled, the circuit in the OP is extremely simple. At least some of the issues have been correctly listed, as well, one being the lack of biasing Vin (to counteract input bias currents) because we were later told that there is an AC coupling capacitor. The hint that this is likely the problem is given by the fact that the circuit seems to work
briefly.