Can you restate that? I'm reading it as one of two different scenarios, either you are describing having both meters in parallel and this jumper you pictured is going from positive of one to negative of another which is going to be a dead short of the power and either blow a fuse somewhere or melt wires, or you don't have them in parallel and this jumper is putting them in series instead, so there is no additional capacity increase, just each meter handing the full load instead of split between them?It can be done. The way to do it is to use Fluke jumper test leads, and first parallel both identical meters. (These special leads can accept leads into them) Then plug the positive working lead into meter 1 (+) and the negative working lead into meter 2 (-).
This keeps power even between both meters.
Then test a 5A load and see how they behave.
I'd have to think you mean the latter but then, there was no point to having two meters except as only a test of the meters themselves, to check the accuracy of one against the other.
As someone already mentioned, what you can do with one of your meters instead is put a low (0.05ohm or lower ideally so the pump doesn't see too low a voltage which will (usually) make it draw less current from that alone unless it is brushless), wattage rated for calculated power dissipation plus some margin) resistor in series, then the multimeter measures the voltage drop across the resistor.
Ohm's law, V=IR, so suppose you had a 0.05Ohm (aka 50mOhm) resistor and measured a 0.6V drop across it, then 0.6= I * 0.05, so I = 12 (amps current). This also allows you to approximate the wattage from 12A * 0.6V = 7.2W, so overshooting that 7.2W and plugging numbers into a supplier site like digikey, makes the following the cheaper for qty one to do the job, though it will need mounted to a heatsink to reach its rated wattage:
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