Considering the equation,
$$E=−L\frac{di}{dt}$$
The negative sign in the above equation indicates that the induced emf opposes the battery's emf.
If we're talking about a purely inductive circuit, the induced emf is equal and opposite to applied emf. Isn't it just like two identical batteries in opposition?
If that's the case, how does the current flow?
How does current flow in a purely inductive circuit if the net voltage is zero?
The problem in this question is that it is based on a completely wrong assumption. This concept of “net voltage” isn’t really a thing. In fact, by Kirchoff’s voltage law your “net voltage” is guaranteed to be zero. So the net voltage being zero does not imply anything about the current.
Isn't it just like two identical batteries in opposition?
No, an inductor is not like a battery. A battery has a voltage that is independent of the current. An inductor has a voltage that is proportional to the change in the current. (A capacitor has a voltage that is proportional to the integral of the current) They are not the same, and having them with opposite voltages does not imply any cancellation of current.
When the source voltage is suddenly made zero then the current will be decreasing at some rate and if an ideal inductor is present in a circuit then that decaying current will cause the magnetic flux to decrease with time through the loop of that inductor.
And In accordance with faradays law if the magnetic flux through a conducting loop is changing with time then there should be an induced EMF in that inductor and that EMF will be induced such that it will support the decaying current (Lenz Law) and this induced EMF causes the current not to decay quickly so, that is why you find the current is still present in the circuit even after the removal of source voltage.
If we're talking about a purely inductive circuit, the induced emf is equal and opposite to applied emf. Isn't it just like two identical batteries in opposition?
If that's the case, how does the current flow?
It isn't, because two real batteries of same emfs acting against each other produce equals emfs whether there is current changing or not changing, and the fact they are real means they have non-zero internal resistance $R$. Then second circuital law from Mr. Kirchhoff states
$$ \mathscr{E} + (-\mathscr{E}) = 2RI $$
which implies zero current, $I=0$.
When a real inductor is connected to a battery, current will flow through the circuit, because there is no static equilibrium like above; the current has to change in time in order for the induced EMF to exist, so initial change of current after the connection is made, is further maintained as the current increases. Because the real inductor also has some internal resistance $R_c$ (let's ignore real inductor's capacitance for now), the second circuital law from Mr. Kirchhoff states
$$ \mathscr{E} - L\frac{dI}{dt} = (R + R_c)I $$ which does not imply $I=0$. In this case, induced EMF does not completely cancel battery's emf, because there needs to be some emf remaining to push the increasing current against the resistance forces in the circuit.
In the case of a perfect inductor and perfect battery, both resistances would be zero and we would get
$$ \mathscr{E} - L\frac{dI}{dt} = 0. $$ which still would not imply $I=0$; instead, current would keep increasing indefinitely.
There is a perfect mechanical analogy. If we push an idle car in a flat ground, we feel a force $$F = -m\frac{dv}{dt}$$ where $m$ is the mass of the car and $v$ its velocity. Of course, at $t=0$, the velocity is zero, but nevertheless its derivative is not zero.
The situation of two opposing bateries is static, like two sumô fighters pushing each other, while staying at the same spot.
