The behavior you're describing (denting stuff) is a function of pressure not force. I could dent a bowling ball with a few grams of force if it was in a small enough area.
I'm not just being pedantic, it's a key to thinking of what you want to do. 24oz of force (yes, it is at a specified current) can be increased over small distances using levers, or you can get a smaller force over a longer distance, again using a lever. Force is just part of the equation; you need to consider the Work that Force will do.
Increasing inductance will affect the solenoid if you are feeding it PWM, but not necessarily in a bad way. PWM works precisely because the inductance averages the applied voltage. For the time that your PWM switch is on, current increases, but at a rate limited by the inductance. When the PWM switch is off, current decreases, but again at a limited rate.
The result of this is that the current (and thus the magnetic flux and force) approximates a steady value that would have resulted had you applied a fraction of the supply voltage according to the PWM duty cycle. For example, if the supply voltage is 12V, and you drive it with a 60% duty cycle, the current you get is essentially the same as you would get by applying a constant \$12V \cdot 0.6 = 7.2V\$ to the solenoid.
Increasing the inductance makes this average better, further reducing the ripple in the current between PWM cycles, or alternately, allows you to use a slower PWM frequency for the same current ripple.
Increasing the inductance also limits the rate of change of current, and thus force. However, it's quite difficult to make the inductance so high that this becomes significant in most applications. If you weren't worrying about it yet, I wouldn't start now.
The bigger problem with adding more turns is usually that the DC resistance of the solenoid also increases. Since the resistance converts electrical energy to heat but doesn't help you create mechanical force, less resistance means higher efficiency. However, more turns allows you to generate more flux with less current. Since resistive losses are given by \$P = I^2 R\$, reducing the current can reduce resistive losses more than reducing the resistance. Yet, more turns also means the induced EMF from the movement of the solenoid will be higher, which requires that you can supply a higher voltage to overcome that EMF to accelerate the mechanical load.
So, you are faced with many trade-offs. In the end, it comes down to optimizing the parameters that are important for your application. Adding more turns isn't good or bad, it's a trade-off. I'd suggest experimenting.
Best Answer
For devices like this, force is proportional to the square of the voltage, minus nonlinearities and losses. So the force you'll get at 40 VDC is (40*40)/(30*30) or about 76% more than 30 VDC, assuming you're already in the zone that overcomes starting friction etc. There's also a return spring -- that has a static force to overcome, so the equation may become: (40*40*kL - kS)/(30*30*kL - kS) kL is the "force of inductor based on squared voltage constant" and kS is "force of spring constant." You can see from this that the higher the kS, the more relative force you'll actually get out at 40 volts than at 30 volts.
That being said, check if your benchtop power supply can be bridged; mine does 30V single, but bridged it can go to 60V.