Lenz's law states that the induced current is in such a direction as to oppose the change producing it.
Back emf and a complete "conducting" circuit will result in an induced current.
The back emf can never exactly equal the applied voltage as then the current would be zero and not changing which would mean that there cannot be an back emf.
So you can think of it as follows.
As soon as a current starts to flow an emf is induced which produces an induced current which tries to oppose that change of current in the circuit produced by the applied voltage.
That induced current slows down the rate at which the current in the circuit increases.
So when deriving equations relating current to time in such a circuit it is convenient to say that at time = 0, when the switch is closed, the current is zero because the applied voltage and the back emf are equal in magnitude.
What you are usually not concerned with is a time scale equal to that whilst the switch is being closed.
Here is an attempt to illustrate the complexity of what happens even as the switch is being closed.
The switch is a capacitor and when open has charges stored on it.
As the switch is being closed the capacitance of the switch increases and so a very small current starts to flow around the circuit as the capacitor charges up.
The change of current is opposed by the induced current produced by the back emf.
As the distance between the contact of the switch continues to get less, the capacitance of the switch continues to increase and a changing current continues to flow.
When the contacts finally close there is already a very small but changing current flowing around the circuit which is being opposed by the induced current produced by the back emf with the back emf slightly less than the applied voltage.
What actually happens during the switch closing process is complicated by the fact that now you have an inductor, resistor and capacitor in the circuit and also that lumped circuit element analysis might be inappropriate to use over such a small time scale?.