Can I use a voltage divider and/or voltage follower to produce 1V from 9V for an op-amp oscillator?
If you want a +/- 1 V wave from a circuit that runs on a 9 V or so supply, then I would recommend that it generate a voltage swinging from 3.5 V to 5.5 V (i.e. 4.5 +/- 1 V).

The trick is what you do with that voltage next. Already I see a red flag because you mention only a positive voltage source but you expect your output voltage to go below 0 V. There are two ways to do this:

1. DC Blocking

In this case, you take the voltage and pass it through an RC high-pass filter (i.e. series C, followed by R to 0 V), then run that to your load. The resistor should be high to keep it from loading down the amplifier, and the capacitor should be big enough to keep the filter corner well below your oscillator's band. R = 150k and C = 1000 pF (NPO/C0G) (f_c = 1061 Hz) should be fine.

2. Redefine Ground

"Ground" is a simplification. There's nothing magic about the node you choose as your reference voltage (0 V). In a lot of applications you want to minimize its impedance, but again, it's still a node that carries current and has a voltage drop if the impedance is high enough.

So in this case, you just redefine your 4.5 V offset voltage as "ground". To lower its impedance, you can even drive it with a unity-gain op-amp. The rest of your circuit treats the 9 V battery as a 4.5 V rail and a -4.5 V rail. This option lets you drive the output directly from the amplifiers.

I'd recommend option 2, but it needs more parts. It's hard to make a better recommendation with so little information.