Electronic – +/-5V to 0…5V and not failing for +/-12V

Positive feedback and negative feedback can be combined to create oscillation (astability). Aside from that, I think you're correct. In the case of the integrator, there's never a step change in the output, so there's never a discontinuity in the capacitor voltage. The only limit is the output voltage range of the op amp.

Assuming an infinite gain with negative feedback implies that the voltages at the two inputs will be equal. Assuming an infinite gain with positive feedback implies that the output will be saturated (unless V+ exactly equals V-).

Even in a real op amp, there's very little current into the input pins (microamps at most). Likewise, the open-loop gain is usually very large at low frequencies (on the order of 10^5 or 10^6 V/V). So the ideal op amp rules are often a pretty good approximation. But all real op amps are limited by their supply voltages, so sometimes that limit is included in an ideal model as well.

The maximum input differential voltage is a breakdown rating on the input stage structure and cannot be ignored; it is in the datasheet for the very good reason that exceeding this may damage the device.

If you power this device from (for instance) +5V and ground, then the output can only slew to within a few millivolts of ground, so if your input goes to -1V, you will definitely be violating the maximum rating of the device.

If you want to convert a bipolar square wave to a unipolar one, a series diode (to prevent the input going negative) and a comparator may serve you better.

Note that the gain bandwidth product of an amplifier is not directly tied to slew rate, which determines what the output looks like. To remain within the small signal region (i.e. the output is not in slew rate limit), the following rule applies:

fmax = Sr/2πVp where fmax is the highest frequency in your signal, Vp is the peak voltage at the output and Sr is the slew rate from the datasheet.