Dunno about the HAI Electric Strike, but yes, the inrush current may be 2 or 10 or 40 times the hold current. 4.5A rather than 450mA.
Not because the coil current is greater when the plunger is outside the coil, as is the case for AC solenoids.
Rather, because the hold-in current is artificially reduced, either by using a separate hold-in coil, or by switching a current limit into the circuit. You do this to get a good pull-in (what you are not getting) but not wasting power and burning out the coil when the door is left unlocked.
A simple tranformer plug pack had no effective limit to current supply: if you shorted it out for long enough, it burned out. A modern switching power supply does have an inherent current limit: if it doesn't burn out it will still only transfer the rated current.
As Mark states in a comment, the solar panel is not large enough. You might be able to buy a converter, that would convert 6.4 A at 17.2 V to 9 A at 12 V. However that is expensive and will only provide operation under ideal conditions. If you buy a second panel, the two working together should be able to power the pump even when the panel is not perfectly lined up with the sun on a perfectly clear day. However, if you do have perfect conditions sometimes, the panels would drive the motor above the rated speed, try to produce more flow then the pump is designed for and overload the motor. It would be better to buy a panel that has a Vmp rating that is only slightly above the motor's full-speed voltage rating and an Imp rating that is comfortably above the motor's full-load current rating. You could connect the motor directly and let the speed vary as the sun moves and the clearness of the day changes.
Two 17.2 Vmp panels with a 12 volt regulator would also be good
Stall Current
The motor's stall current is the current that it draws at zero speed regardless of the torque that it is loaded with. That current is determined by the applied voltage and the winding resistance. If the power supply limits the current, the torque that the motor can produce is limited by about the same percentage. With a centrifugal pump, very little torque is required to get it moving. The only resistance to rotation is the bearing friction in the pump plus the bearing and brush friction in the motor. The rotor of the pump is just stirring the water a little, not producing any pressure or flow. As speed increases, torque increases in proportion to the square of the speed as the pump builds up pressure. If the current is limited to 50%, the pump can accelerate until the pump requires about 50% of rated torque or about 0.5 X 0.5 = 0.25 or 25% of rated speed. If more current is available, the motor will draw more and accelerate faster, but it is totally unnecessary to make more current available than is required to run the pump at full speed. The friction torque at at standstill may be a little more than friction with the shaft turning, but no where near the total torque at full load.
Best Answer
It's probably adequate. The pump draws 60W continuous, which is 5A, according to the nameplate, and you are proposing a 10A power supply.
Diaphragm pumps, to my knowledge, do not tend to draw enormous current on start-up, and the power supply, if the ratings are real, is capable of supplying some startup surge, so a 2:1 factor is probably okay.
The only way to be sure, really, is to test it. You could also go for a much bigger power supply to be more sure.