Confused with turn on time definition of an SCR
The SCR is on when the voltage across the SCR falls to a low voltage and current is flowing. How else could you get current to rise up to the "holding current" ?

If your circuit takes a long time to establish the holding (latching) current, then yes, the SCR will turn back off when you remove the gate signal. 

Typical turn on time for an SCR is 1 microsecond. Apparently your SCR is specified as 5 microseconds.

Turn on time also includes what is called the "spreading time" which is the time for the SCR to tun on across it's entire surface. 

      EDIT 1 : 


Turn on time of SCR :  includes three small intervals as delay time (td) rise time(tr), spread time(ts).

Rise time : Rise time of SCR in the time taken by the anode current to rise from 10% to 90% of its final value. At the same time anode voltage will fall from 90% to 10% of its initial value Va.

Delay time :  After application of gate current, the thyristor will start conducting over a very tiny region. Delay time of SCR can be defined as the time taken by the gate current to increase from 90% to 100% of its final value Ig. 

Spread time : It is the time taken by the anode current to rise from 90% to 100% of its final value. At the same time the anode voltage decreases from 10% of its initial value to smallest possible value. In this interval of time conduction spreads all over the area of cathode and the SCR will go to fully ON State. Spread time of SCR depends upon the cross-sectional area of cathode.

In the case you show above, the calculation shows how much time is required for anode current to rise to the minimum latch current. And the gate would need to be energized for that amount of time (minimum) to keep the SCR on.