Patched example of non-regular language that makes it through the pumping lemma

Let L = {aⁱbⁿaⁿ : i ≥ 1, n ≥ 0} ∪ b^*a^*

Let p = 1. Then suppose w is any string in L with length ≥ 1.

Let x = ε, y = first character of w, z = the rest of w. Then w = xyz, |xy| ≤ 1, and |y| ≥ 1. Moreover, xyⁱz is in L for any integer i ≥ 0.

So L satisfies the pumping lemma. But it's not regular.

Suppose that L were regular. Then L' = L ∩ abb^*a^* would also be regular. L' is just {abⁿaⁿ: n ≥ 1} which you can easily show to be non-regular using the pumping lemma.