Electronic – Why does this single supply non inverting Op-Amp saturate

Like markrages says (and like you did in your second schematic) you need to bias the non-inverting input. The document says that

"the input impedance = R1||R2 for minimum error due to input bias current"

That's how it should be, but they don't do it! If you leave it floating it will assume \$GND\$ or \$V_{CC}\$, depending on the opamp's design. The 100k\$\Omega\$ value you use is not the right one, though. For optimal bias this should be equal to 10k\$\Omega\$||12k\$\Omega\$ = 5.6k\$\Omega\$. This will affect the frequency response, since your input is now a high-pass RC filter, with

\$ F_C = \dfrac{1}{2 \cdot \pi \cdot R \cdot C} \$

This will also be the response of your amplifier if you add a capacitor in the feedback path:

\$R\$ goes to \$V_{CC}/2\$ instead of \$GND\$.

From a DC point of view R1 and C1 aren't present, so the output bias will be the same as the input bias (voltage follower). This way of biasing is preferable over connecting R1 to \$V_{CC}/2\$, because you would have to take the voltage divider resistors into account to calculate R1. That's unless you use a "hard" \$V_{CC}/2\$ with a low impedance, like from a voltage regulator.

Same goes for R. If you do want to use a voltage divider you can use resistor values equal to 2 \$\cdot\$ R, then you can eliminate R altogether.

Equations:

\$ F_C = \dfrac{1}{2 \cdot \pi \cdot R1 \cdot C1} = \dfrac{1}{2 \cdot \pi \cdot R \cdot C} \$

Yes, as you suspect the CA3130EZ is not unity gain stable without compensation capacitor(s). On page 3 of the datasheet states that a capacitor (called \$C_c\$) of at least 47pF be placed from pin 1 to pin 8 of the device. \$C_c\$ provides Miller compensation, and reduces bandwidth of the part. Circuits shown with unity gain also typically have a 2kOhm || 0.1uF feedback network, and a 25pF load. See the Bode plot of open loop response on page 14 for details of effects of various values of \$C_c\$ and \$C_L\$.