Question about transient behaviour in op-amp circuit
Obviously, this cannot occur in a real circuit


You have several problems. 

You are using derivatives on something that is discontinuous. That usually tends not to work. Also, the opamp's output voltage will go to infinity at t=0. I will now explain...


  The switch closes at \$t=0\$. The voltage source is \$V_0e^{-t/t_0}\$.


So, after the two rather useless resistors which divide this by 2, when \$ t=0^- \$ which is a tiny instant before t=0, we have \$ V= 0 \$. Then at \$ t=0 \$ the switch closes and we have \$V=V_0/2\$, thus V is discontinuous.

Our nice perfect opamp must adjust its output voltage to keep its two inputs at the same potential. So, at t=0 the output of the opamp will go up. However, there is a perfect inductor in the feedback loop. And the current in an inductor cannot change instantly. And the opamp's negative input voltage V is \$ R i_L \$. And at t=0, \$ i_L =0 \$.

Thus, the voltage at the two inputs of the opamp is not equal. IN+ is at Vo/2 and IN- is at 0V. So, the perfect opamp model stops working, but you got other problems. For the ideal opamp, there is only one choice...  its output voltage instantly jumps to +infinity. 

Now, at t=0 voltage across the inductor which is \$  V_{inductor} = L\frac{di_L}{dt}\$ is thus +infinity. 

If we break enough math, spill the pieces over the floor, then roll over it forward and reverse with a tank, which you did, we can thus integrate that and conclude that \$  \int V_{inductor} = L i_L\$ and therefore right after t=0, give or take, more or less, \$ i_L = 1/L \int +\infty = V_o/2R\$ ...

Tada! Done.

Now, your problem is you didn't notice this, so your you got the wrong result in "Approach 1". No blame, we all make mistakes, and the ideal opamp model really invites mistakes too, since any deviation from ideal operating point creates impossible conditions.

I didn't check the Laplace Transform calculations, but let's examine the circuit again. Let's throw away the switch and both resistors, and use the "V" node as input. We now have a bog-standard non-inverting amplifier. Its gain is:

\$ G = 1+ Ls/R \$

Now, as is obvious from this equation, G goes to infinity as frequency goes up. This kinda works like a differentiator. But it is given a discontinuous signal as an input. Thus it derivates a non-derivable input. Hence your Laplace results have a delta.

At infinitely high frequency, the inductor's impedance is infinite, thus it is an open circuit, and we can remove it from the schematic: on discontinuous inputs, the opamp no longer has feedback, so the ideal opamp model cannot apply.

In real life, opamps are not infinitely fast, therefore the phase lag caused by the inductor in the feedback loop will turn the opamp into an oscillator.

So this circuit is a trap ;)