How to bias an AC signal correctly?
You are passing what appears to be 60 Hz through a 100 nF capacitor. That capacitor will have an impedance of: -

$$|X_C| = \dfrac{1}{2\pi f C}$$

Plug in the numbers and you get 26.5 kohm so, if you need very little attenuation of the AC signal the resistor value needs to be many, many times the value of 26.5 kohm.

In your last scenario the equivalent resistance is 680/2 kohm and this is much bigger than 26.5 kohm hence, it works (somewhat). In your early experiments you didn't have a resistor at all i.e. the capacitor output fed straight to Vref and there is no guarantee that Vref will have anything other than a few ohms to a few hundred ohms impedance effective series impedance.