Understanding LDO Dropout
Alright so I think I found a clear explanation for that statement. According to this paper one test method is to:

measure the output voltage   when   VIN   is   1   V   above   the   nominal   output  voltage  and  then  load  the  LDO  with  the  specified  test  current.  The  input  voltage  is  then  reduced  until  VOUT  drops  by  a  specified  number  of  millivolts.  For  example,  4.3  V  is  applied  to  a  3.3-V   LDO.   The   output   voltage   is   measured   (assuming at 3.3 V), and then the input voltage is reduced  until  the  output  voltage  measures  3.2  V.  Dropout is defined as the input voltage minus the output  voltage  at  this  point.  If  VIN  =  3.5  V  when  VOUT  =  3.2  V,  the  LDO  has  a  300-mV  dropout  voltage. An LDO tested in this method will have a note similar to that shown below.
“Dropout     voltage     is     defined     as     the     differential  voltage  between  VOUT and  VIN  when  VOUT  drops  100  mV  below  the  value  measured  with VIN = VOUT + 1 V “

In the datasheet I was reading the manufacturer opted to use 2% or 66mv in the case of 3.3V as opposed to the 100mv in the example above.
As for my specific application, I replaced the regulator with one rated for 500mA. The new one is working perfectly. I will wait for my current probe to see using my scope how much exactly is the ESP spiking. It seems like such a quick spike over 250mA that my MM's Max function can't catch it.