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?.
If there is no resistance in the circuit, the rate of change of current in the circuit does not change, as explained here: inductor back EMF.
Thus, the decrease in the magnitude of the rate of change of current as time progresses must relate to there being resistance in the circuit.
In the circuit, you have $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$.
As time progresses, the current in the circuit increases, so $V_{\rm resistance}$ increases. Because $\mathcal E_{\rm battery}$ stays the same, $\mathcal E_{\rm back} $ must decrease, as does the magnitude of $\frac {dI}{dt}$.
To my knowledge, if there is a back emf, it should oppose the applied emf and the sum of these "emfs" should result in a net emf. Current should only increase if the net emf increases... No?
To understand what is happening one needs to consider where the equation $\mathcal E_{\rm Battery} - V_{\rm resistance} = \mathcal E_{\rm back} = (-)L\frac {dI}{dt}$ come from and the use of the terms potential difference and emf.
From what you have written I assume that you have never queried as to what happens in a simple circuit which consists of an ideal battery and a resistor.
In such a situation the potential difference across the battery $V_{\rm battery}$ is equal in magnitude to the potential difference across the resistor $V_{\rm resistor}$ so that $V_{\rm battery}-V_{\rm resistor} =0$.
If the equation is multiplied by the current $i$ then the equation becomes $V_{\rm battery}i-V_{\rm resistor}i =0$ and it perhaps now becomes apparent that the equation is a restatement of the law of conservation of energy for electrical circuits, electrical power output from the battery is equal to electrical power dissipated in the resistor.
The only thing to add is that $V_{\rm battery}$ is also called the emf of an ideal battery, $\mathcal E_{\rm battery}$.
A slightly more complex example is an ideal battery used to power an ideal electric motor.
The steady state condition is that there is no current in the circuit because the emf of the battery, $\mathcal E_{\rm battery}$, is equal in magnitude to the back emf produced by the motor coil rotating in a magnetic field $\mathcal E_{\rm back,motor}$ thus $\mathcal E_{\rm battery} - \mathcal E_{\rm back,motor}=0$.
Now suppose that the motor is made to do some work by you pinching the axle of the motor.
The speed of revolution of the motor decreases and hence the back emf produced by the rotating coil, $\mathcal E_{\rm back,motor,load}$, decreases and what is the net input power, $\mathcal E_{\rm battery}i_{\rm load}-\mathcal E_{\rm back,motor,load}i_{\rm load}$, equal to?
In this case is is equal to the mechanical power output from the motor, the work done per second against the frictional force that you have applied to the axle.
Returning to your question one has to consider a dynamic situation.
Points to note are that connecting a battery to a length of wire will results in a current flowing in the wire.
If the circuit has inductance an emf will be induced because of the changing current (Faraday) and that induced emf will oppose the change producing it (Lenz) which in this case is the changing current.
So connect an ideal battery to and ideal (no resistance) inductor.
At the start even with no current flowing the current must change otherwise there would be no back emf from the inductor and then there is an unnatural situation with a battery being connected to a piece of wire with no resistance and no current is flowing.
Now we come to the situation which you are unsure about.
Two equal and opposing emf, $\mathcal E_{\rm battery}$ and $\mathcal E_{\rm back}$, and yet the current changes and noting that if there is no change in the current $\mathcal E_{\rm back}=0$.
$\mathcal E_{\rm battery}i-\mathcal E_{\rm back}i$ relates two powers.
$\mathcal E_{\rm battery}i$ is the instantaneous power delivered by the battery.
So what is $\mathcal E_{\rm back}i$?
It is the rate of change of the magnetic energy stored in the inductor.
So in this situation the instantaneous electrical power delivered by the battery is equal to the instantaneous rate of increase in magnetic magnetic energy stored by the inductor.
In a time $\Delta t$ the battery delivers electrical energy equal to $\mathcal E_{\rm battery}i\Delta t$.
During that time the magnetic energy stored by the inductor changes by $\frac 12 L(i+\Delta i)^2 - \frac 12 Li^2 = Li\Delta i + \mathcal O \Delta i^2$ and you will note that $Li\Delta i = L \frac{\Delta i}{\Delta t} i \Delta t= \mathcal E_{\rm back} i \Delta t \,(=\mathcal E_{\rm battery}i\Delta t)$.
Perhaps do not think of the emfs as battling it out for domination rather that one emf is the source of electrical energy and the other emf is the sink (user) of electrical energy and they are equal because energy is a conserved quantity.
Best Answer
You are right that right when we close the switch the voltage across the inductor is equal to the applied voltage. However, you are misinterpreting what a potential difference of magnitude $v=L\cdot\text di/\text dt$ means. This equation doesn't say if there is a potential difference across the inductor then there is current through the inductor. What it says is that a potential difference across the inductor is associated with a change in current through the inductor. Therefore, since the voltage across the inductor is non-zero at $t=0$, we know the current is changing at $t=0$.
The current increases like $$i=i_0\left(1-e^{-t/\tau}\right)$$ So it is increasing, and there is an exponential function, but usually "increasing exponentially" means it keeps growing and growing more rapidly without bound. This is not what is happening here.
As the current in the circuit increases the voltage across the resistor increases. Therefore, the voltage across the inductor decreases. Based on our previous discussion, this means that the change in current must be decreasing. Hence this "voltage trade-off" happens at a slower and slower rate. This causes the current to approach a steady value where the increase over time decays exponentially.
Keep in mind that "oppose" does not mean "block".
It seems like your confusion stems from what we discussed initially. You are mixing up the current and its derivative. The voltage across the inductor tells you nothing about the current in general. It tells you how the current is changing.
Also, you say that you understand things from the equations, but I would argue that if you don't understand how the equations model reality then you haven't truly understood the equations. It would help for you to look at how the equations are derived. Make sure you understand the physical significance and motivation for each step, each equation, etc. This is an important step in the learning process, so I will leave that job to you.
I hope this answer is a good scaffold to hold up the deeper understanding you will develop here.