What will be the result of a step input?
This circuit is actually a common configuration and the only factor that cannot be determined due to the ideal op amp is the eventual high-frequency roll-off.

It's perhaps easier to recognize when redrawn like this:



I've used a TL071 as the op amp just because it's easier for me than getting an ideal op amp in play and used \$R_f\$ (for feedback resistor) instead of \$R_1\$ and \$R_g\$ (for gain resistor) instead of \$R_2\$ because I find this makes it easier to make sense of the gain expressions when they get complex.

This circuit produces the following frequency characteristic:



The right-most gain roll-off after 1 MHz is due to the real frequency response of the TL071, so would not be present using an ideal op amp.

However, the single pole, single zero characteristic would appear just as it does here at roughly 1.6 kHz and 3.2 kHz respectively.

The pole appears at the frequency where \$X_c = R_f\$ which is:

$$ \frac{1}{2 \pi R_f C} = \frac{1}{2 \pi 100 \cdot 1e-6} \approx 1592 Hz  $$

The zero appears at twice that frequency.

For a 1V step, this produces a response that looks to me like the result you proposed:



I haven't verified that you got precisely the right time constant and I'm not sure I would have chosen the equivalent circuit you did (it seems to neglect the effects of feedback). But you seem to have arrived at the correct qualitative behavior, if not the exact behavior.

The additional gain and bandwidth of an ideal op amp would not change the character of this step response.

The transfer function for this circuit, in "low-entropy" form is:

$$ \frac{V_{out}}{V_{in}} = \frac{R_f+R_g}{R_g} \frac{s(R_f\parallel R_g)C + 1}{sR_fC+1} $$

Substituting values for this particular circuit:

$$ \frac{V_{out}}{V_{in}} = \frac{100+100}{100} \frac{s(100\parallel 100)(.000001) + 1}{s(100)(.000001) +1} = 2\cdot \frac{.00005s + 1}{.0001s +1} $$

shows us the DC gain (2), the single pole at \$\omega_p = 1/.0001 = 10,000\$ and the single zero at twice that frequency \$\omega_z = 1/.00005 = 20,000\$. In Hertz these are 1591 Hz and 3183 Hz respectively.