Capacitor loop with no resistance
Guesstimate the resistances you would have in the real circuit, and insert resistors in the sim circuit to represent them.

A little more explanation of why the simulator reports an error without them may be useful : the instantaneous current when the battery and capacitor are connected is independent of the capacitance; it is purely the voltage divided by the resistances in series. Which, with ideal components, is the well known "divide by zero " error.

The capacitance will have some value of "ESR" = equivalent series resistance, which you may find from datasheets. (Some datasheets model ESR as loss, or "tan delta" instead) This page allows you to select capacitors by ESR; values from 0.003 ohms to 1.6 kilohm with 0.015 to a few ohms being common.

The battery will also have an ESR which will increase as the battery drains; again, see the datasheet. A fully charged NiCd or NiMh may have 0.1 ohm or more ESR, up to tens of ohms for a small button cell.

For the wiring or PCB traces, you can find  the resistance in ohms per meter for approximately the right wire, (e.g. 0.1 ohm/m for 26SWG) and estimate the length of wire required (say 10cm, or 0.01 ohm).

Any one of these will break the loop and produce a circuit that can be simulated; modelling all of them will produce a more accurate simulation (may not change the results very much)

There may also be better simulation models for components such as capacitors, that allow you to set the ESR (and other parameters) and other features of real components.