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I just happened to re-read this question, and I now think I see better what Microchip was trying to say here:

<img src="https://electrical.codidact.com/uploads/1dNehfVbWqpX8kovK6iks5wQ">

On closer examination, I think this is actually correct but badly worded.  I believe it should be interpreted as:

<blockquote>Dropout voltage is defined as the input-to-output differential when the output voltage drops to 99% of its normal value, <b>compared to when</b> V<sub>OUT</sub> + 1V is applied to V<sub>IN-</sub>.</blockquote>