Electronic – Op-amp differential amplifier circuit for inputs greater than positive rail

Feed back will try and make the two inputs the same. So the inverting input is also at zero volts. The negative voltage (CV-) is sucking 5/3k = 1.66mA from that node. The only place that current can come from is the opamp output. You need 1.66 mA flowing through R3... and that gives 1.66V on the opamp output. (No current can come from R5 because both sides are at zero volts... that is only true when Vin is zero.)

Okay so you seem to be under some confusion about what that charger was used for. The thread you linked from stack-exchange did not say that this schematic was compatible with Li-ion or Li-poly, just that it was very accurate. Now to answer your questions in order.

1. What ESD Diodes? That op amp doesn't have any in it's schematic nor is it normal for BJT input op amps as they tend to be quite resilient (up to 5V reverse and sometimes up to 100V forward, then they poof). So unless you're providing some that I don't see, there's nothing there to clamp.
2. If you still want to make this with a 5V supply, then you just need to change the resistor values to move them into common mode range. I'd also select an op amp that has its common mode range up to Vcc and some voltage above ground (Vcm = [Vcc] -> [Vss + 2V]) instead of from ground to Vcc minus some voltage (Vcm = [Vcc - 2V] -> [Vss]). After all, your cell will die around 2.5V so what's the point of having common mode input range below that?
3. It is possible that that schematic is wrong. That would make the most sense, but just because it looks like the circuit shouldn't work doesn't mean that it can't. Remember that transistors are odd little ducks and we take great liberty when teaching about them and talking about them so that the person listening doesn't need a PhD to understand them. At inputs of around 8.4V to both the negative and positive terminal of the op amp, those input PNP transistors should be well into cut-off. But cut-off doesn't mean "no current" it just means "less current than we could possibly care about." Except when we have 100dB of gain. That's the kicker. Even though both of those transistors are "off"; one of them is still slightly unbalanced from the other (8.4V vs 8.3998V from my spice testing). That means that one transistors is slightly less cut off than the other and that a slight bit more current is flowing through its cut-off mode. From there it's all op-amp internals (that current gets converted to voltage by the transistor load which then gets amplified further until it pops out the output). These transistors, assuming they're made in silicon, would still be operable since most silicon BJTs (that I know of, please tell me if I'm wrong) can withstand up to 5V reverse bias between the base and emitter (8.4V-5V = 3.4V, close but still no magic smoke). So, this shouldn't work, but it totally could work, if you're lucky. Remember that since this is outside spec that you won't get any guarantees from anyone (myself included, one bad fluctuation away from poof is a bad design. I do not condone or support the murder of op-amps in the name of science or otherwise. I also don't condone the design).

I look forward to the results.