OpAmp with common-mode voltages and zero feedback current
Actually with finite gain [...] the argument would be even more compelling - finite open loop gain multiplied by zero potential must be zero!

I find it even more clear to see with finite gain. You are forcing only one input to 1V. The other depends on the voltage source and the op. amp. output for its voltage. If you break the negative feedback, you remove the mechanism that tries to &quot;zero&quot; the differential input voltage.
This circuit uses an op. amp. only with finite gain as a DC non ideal characteristic (in fact, a very small one, 1000, to make it even more clear):

There is current through the resistors in this condition.
$$\frac{1-0.99901088}{1000} = 989.1 nA$$
$$\frac{0.99901088-0.98911969}{10000} = 989.1 nA$$
If you break the feedback with this ideal model, except for the finite gain, everything works as you expected (\$(1 - 1) \times 1000 = 0\$):

Trying to follow a thought process while multiplying \$0 \times \infty\$ is really a problem. As the open loop gain tends to infinity, the differential input voltage tends to zero (with negative feedback). The current through the resistors will tend to zero, but non-zero current means that \$ V_{in-}\$ &gt; Input- pin \$&gt; V_{out}\$. The two op. amp. pins will tend to the 1 V @ the non-inverting pin, but not together if the feedback resistor is non-zero.
Nice reading here: https://math.stackexchange.com/questions/28940/why-is-infty-cdot-0-not-clearly-equal-to-0.