OK, here is the modified solution:

1. Still draw small rectangle (2a+1) x 2b, and assume (2a+1) is the height.

2. For the outsiders, if the height is even number, the solution is the same as before.

3. If the height is odd, break the very top (or bottom) stripe first, which it's width is even and can be divided. Then it's same as before.

4. If there's no extra stripes on top or bottom of the small rectangle, then

(a) The left or right stripes on sides of the the small rectangle have to be both even number or both odd number. If both are even, simple...

(b) If both are odd, strip down to a single extra stripe, and also reduce the small rectangle down to 2x3 as before. In the end, instead of 2x3, it should be 4x3, as I'll show you below:

XOYX

XYYX

XYOX

where "O"s are the missing pieces, "O"s and "Y"s made the small rectangle and "X"s belong to the extra single stripes on both sides. Now this formation can be divided as follows (same letter in one piece):

AOBB

ACCD

EEOD

The explanation is too lengthy. Sorry.