Reducing IO Expander current consumption when driving LEDs
The link provided in the question does not point to a port expander. This link http://www.nxp.com/docs/en/data-sheet/PCA9538.pdf points to a port expander with the same strategies shown in the question.
This datasheet does not refer to leakage current, but, correctly, to IDD.
From the datasheet:

Minimizing IDD when the I/Os are used to control LEDs
When the I/Os are used to control LEDs, they are normally connected to VDD
through a resistor as shown in Figure 11. Since the LED acts as a diode,
when the LED is off the I/O VI is about 1.2 V less than VDD. The supply current, IDD, increases as VI becomes lower than VDD.
Designs needing to minimize current consumption, such as battery power applications, should consider maintaining the I/O pins greater than or equal to VDD when the LED is off.

Here's why:
Each pin of your port expander has an output buffer (which can be disabled, e.g. during reset or power down, if any) and an input buffer, which cannot be disconnected. That input buffer is typically an inverter.
If either the port expander is in the reset/power down state, or the pin is configured as input, the voltage at the GPIO pin would be something about Vdd-Vled (of course - Ileak*R). Assuming Vled = 2V 1.2V (EDIT: The datasheet states 1.2V and not 2V, most likely because the current will be extremely low: the input leakage current), the input of the inverter will be extremely close to the logic threshold (VDD/2), and both MOSFETs will be on (direct path conduction). The 100-k resistor pulls up the input, forcing Vdd to the inverter, turning off the pMOSFET.
Similarly, the arrangement on the right, you'll have 5V-2V = 3 V, which is high enough to turn off the pMOSFET.



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