It depends on the application. Assuming that the internal resistance is 0 will introduce some error in your circuit's output. Whether or not that is OK depends on the size of the resulting error, the size of other errors, and the required accuracy for the application.
is it really ok to assume that internal resistance of a cell is absent in a circuit
Sometimes, not always.
Consider a small 9V battery like this http://www.westwealdfalconry.co.uk/_productimages/Duracell.jpg If that battery is powering just a small resistor & LED then the current draw will be low (perhaps 10mA). There won't be much voltage drop due to the internal resistance of the cell so the battery voltage will be close to 9V.
On the other hand if you tried to use the same battery to power a 50W light bulb the current required would be large (50/9 = 5.6A). The voltage drop due to the internal resistance would be very large. So large that the voltage wouldn't actually be anywhere near 9V (edit: You would have to factor in the internal resistance to calculate the actual current because 50/9 would no longer be accurate enough).
The internal resistance of a battery can also change so that when new (fully charged) it may deliver 9V on load, but then as the battery discharges the resistance increases so that when nearly flat it only produces 9V off load. As soon as you connect the load the voltage falls.