I'll tidy this up "later".
Duty calls. This provides enough guidance.

With solenoid operated current will be due to inductance + coil resistance. Current at 50 Hz will be greater than at 60 Hz due to lower impedance of inductor. This can be as much as 60/50 = 20% higher but will be somewhat less than this due to the coil resistance. Main losses will probably be I^2R losses in coil resistance and these will be up to 40% higher due to I^2 term. 

Extra thermal losses may be OK depending on ambient temperature, enclosure and manufacturers ratings. Usually I'd expect it to be "probably OK but it may not be. look at resistance and rated operating current - which will show you what the effective resistance is to AC so how much the inductor is affecting things.   
eg made up example.  

Voperating = 220 VAC.  
Ioperating = 100 mA.  
Rcoil =   1000 Ohms.  
So effective "AC resistance" = V/R = 220/0.1A = 2200 Ohms.  
R + L impedances will vector sum ()Zcoil = sqrt(R^2+Zl^2) but usually close enough Zcoild = R + Zl (calculate actual values if desired)    
So simple way Zl ~= 2200 - 1000 = 1200 Ohms.    
So on 50 Hz Zl will be ~= 1200 x 50/60 = 1000 Ohms.  
Overall Zcoil will then be R + Zl = 1000 + 1000 = 2000 Ohms.  
So current will rise by 2200/2000 = +10%.
Heating in Rcoil will rise 20%.

Let's try "proper" calculation above.

Reffective =220 Ohm.  
Rl = 1000  
sqrt(Zl^2 + 1000^2) = 2200    
Zl = (sqrt 2200^2 -1000^2) = 1960 Ohms.  
So above figure of Zl = 1200 Ohms was "too far off" :-(    

So Zl 60Hz = 1960.  
Zl 50 Hz = 50/60 x 1960 = 1633 Ohms.  
Rcoil_AC = aqrt(1633^2 + 1000^2) = 1914 Ohms AC.  
I increas = 2200/1914 = 1.15x = 15%.  
Power increase = 32%.  
Prior power in Rl = 1000 x 0.1A^2 =  10 Watts.   
New power in Rl ~= 13 Watts.  
Q: Does it matter?  
A: Probably not, but, it may.