Isolating GPIO’s using diodes

Thus, when the switch is closed the digital value of the GPIO changes from high to low, and when open there is only a small trickle of current which saves battery life immensely.

You have the wrong idea. A pull up should not result in much current aside from the leak current of the gpio

^{simulate this circuit – Schematic created using CircuitLab}

In the first picture, a pull up pulls the GPIO IN high. The only current is the leakage current required to measure the voltage input level. This is a few nA, a fraction of a milliamp. This can most often be checked as Input Logic High/Low (ILH, ILL) in a datasheet but I can't find it in your product. Assume 10nA or less on average.

It's only in the second picture that a significant current can be drawn. When the button is pressed, there is a straight path from V+ to Gnd, through R2. Assuming 3.3V, that's 3.3V / 47000Ω = 0.00007A It's 70000 nA. Which looks large in comparison, but that's still only 0.07mA or 70µA.

The significant current draw is only when the button is pressed. So simply design your circuit so the default state of the button is where the button is open.

To reverse your logic you just need to reverse the pull up and make it a pull down.

^{simulate this circuit – Schematic created using CircuitLab}