Think of wire as a low-value resistor. The thicker the wire, the less its resistance. The question then becomes "Is adding a resistor to the load going to cause a problem?" The voltage drop across the resistor will increase proportionally with its resistance. If you think of it this way, the goal is to reduce resistance to a usable (and safe) point.
You can look up wire resistance values on various wire gauge charts and calculators like this one. Network cable is ~24 AWG (you didn't specify what type of network cable). 24 AWG has a resistance of about 84 mΩ per meter. If your cable length is 50 meters, then you're adding 100 meters (both supply and return), or 8.4Ω to the circuit.
Treating the solenoid as a fixed resistance of 65Ω: at 24 V the total current will be limited to 327 mA instead of 370 mA, and the voltage drop by the wire will be about 2.75 V. In that case, the solenoid will "see" 21.3 volts instead of 24.
If you use two conductors instead of one, you'll halve the resistance. Total current would be 347 mA, voltage drop on the wire 1.46 V, and the solenoid "sees" 22.5 V.
You'll need to test to see if either of these configurations allows the solenoid to operate properly. As a valve or any mechanical thing ages/oxidizes/collects debris, what works now may not work later.
Andy Aka wrote a great answer discussing inrush current on AC-powered solenoids. The figures I used above are based on the initial state as you mentioned in the question. As the solenoid settles, current should decrease a bit, which means the values I calculated should be "worst case."