Series inductor as discontinuity in transmission line
... undesirable 2 factor.  I would like that \$ V^{++} \to V^+ \$ for \$ t \to \infty \$, but when \$ I_L(t) \to V_0 / (2Z_0) \$ there is an unavoidable voltage divider, maybe due to the circuit.


The factor of 2 has nothing to do with the inductive discontinuity. To see that this is true, take the limit as \$L\to{}0\$, and you'll still have the factor of 2. 

The factor of two is fundamental to using matched sources and loads with transmission lines. If you want to have a matched source generate a signal on a transmission line with amplitude \$V\$, you need the amplitude of that source to be \$2V\$.


  Is it possible to cancel this \$ 2 \$ factor (and obtain exactly \$ V^{++} (t) = V^+(1 - \exp{(-t/\tau_L)}) \$ like in the [capacitor][1])?


No. The factor of 2 was also there with the capacitive discontinuity, if you did your math right.