One could "break cycles" in order to flip edges or twist corners, but (I think?) most people just use algs for that, so assuming that, we can ignore the orientation of the pieces entirely. In other words, we're asking for the probability that we have only one corner cycle and only one edge cycle. (Please correct me if this is wrong and not what you're asking for, because I have no idea how people do BLD.)

Consider the two cases based on permutation parity. Suppose the corner/edge permutation parities are even. (They must be either both even or both odd.)

P(one corner cycle | even parity) = (P(corner 7-cycle) + P(corner 5-cycle) + P(corner 3-cycle) + P(corners solved)) / P(even parity)

= [(C(8,7)6! + C(8,5)4! + C(8,3)2! + C(8,1)0!)/8!] ⋅ 2

= [1/(7⋅1!) + 1/(5⋅3!) + 1/(3⋅5!) + 1/(1⋅7!)] ⋅ 2

= 43/120

By the same reasoning:

P(one edge cycle | even parity) = [1/(11⋅1!) + 1/(9⋅3!) + 1/(7⋅5!) + 1/(5⋅7!) + 1/(3⋅9!) + 1/(1⋅11!)] ⋅ 2

= 48187/217728

As the corners and edges are conditionally independent once we know the parity, we can conclude that the probability of not having to break cycles when the parities are even is:

P(one corner cycle, one edge cycle | even parity) = P(one corner cycle | even parity) P(one edge cycle | even parity)

= 2072041/26127360

≈ 7.93%

Then we can do the same computation for the odd parity case:

P(one corner cycle, one edge cycle | odd parity) = 39784637/326592000

≈ 12.18%

And conclude:

P(one corner cycle, one edge cycle) = (1/2)(2072041/26127360) + (1/2)(39784637/326592000)

= 131370299/1306368000

≈ 10.06%