Can this circuit be a Schmitt trigger?
Short: 


Yes, it's a Schmitt trigger.
Yes, you have made a mistake.
Output low / high happens at Vin =~ -10V*  
Output high / low happens at Vin =~ -15V*.


*I have slightly adjusted Vout range to make sums easier. Result is essentially the same. As shown the zener is being used in an unapproved manner. 



Longer:  Loosely a Schmit trigger can be considered to be a comparator with hysteresis or positive feedback such that the switching decision point alters when the circuit ir triggered to provide a "deadband". On that basis this IS a Schmitt trigger circuit.

Using a zener to clamp Vout as shown is a VERY bad idea.
The opamp is using all its output to try to drive the zener high. This is a non-approved way to do what is desired BUT OK for working through the example 

Without using your notes.
 Vout swings from about -0.8V to about 4.3V. For practical purposes we can work with 0V and 5V and get an OK answer to the main question.

Vref on the inverting input (OA-)is -5V. This does not change.

Vout is low = 0V when the non inverting input OA+ is below -5V and high = +5V when the non inverting input is above -5V.

When Vout = 0V then the trigger condition occurs when Vin = -10V, so OA+ = -5V.

When OA+ rises above -10V Vout rises to +5V.
 So immediately OA+ rises to -2.5V (half way between Vin at -10V and Vout at +5V.
So the opamp is now ell away from the switching point.  

To switch back OA+ needs to fall to -5V and to achieve this Vin will need to fall to -15V (as OA- is now +5 - (5 - - 15V) = 5 - 10 = -5V.)

SO the switching boundaries are
 -10V rising = output low to high and
 -15V falling = output = high to low.