Triangles $DAE$, $DCE$ and $DBE$ form a quadrangle $ABCD$, where $\angle BAD$ and $\angle BCD$ are right angles.
I have a scenario in which I want to find the angles $\angle ABD$ and $\angle DBE$, which form the rotation of BE from BA, in terms of $\angle ABC$, $\angle DCE$, $\angle DAE$ and $DB$. I got as far as
$$\angle DBE = \tan^{-1} \frac{DA \tan (\angle DAE)} {DB}$$
but am unsure about how to define $DA$ in terms of $\angle ABC$, $\angle DCE$, $\angle DAE$ and/or $DB$. Or maybe this is unsolvable for the terms that I want?
I stumbled upon the Rodrigues' rotation formula and while it sounds like it might be the way to go about this, the breakdown of it completely lost me.
This is all to be used in a script to rotate an extruded shape that has been multiplied through a corner.
Following from Cosmas Zachos, $$\angle ABD = tan^{-1}(\frac{tan(\angle ABC)sin(\angle ABC)tan(\angle DCE)} {sin(\angle ABC)tan(\angle DCE)+tan(\angle DAE)tan(\angle ABC)})$$
where $\angle ABC \ne 90$ ?
$$\Longrightarrow\angle DBE = tan^{-1}(\frac {BDsin(\angle ABD)tan(\angle DAE)}{BD})$$
Your proposed rotation is quite misconceived, when plain Alexandrian trigonometry will work. You started out right.
Try to determine $\theta \equiv \angle ABD$ first, through $\triangle ABD$ , then $\triangle ADE$; but also $\triangle BCD$, then $\triangle CDE$, both depending on it differently, $$ DA=DB \sin \theta \Longrightarrow DE= DA \tan ( \angle DAE ) = DB \tan ( \angle DAE ) \sin\theta \\ CD=DB \sin(\angle ABC-\theta) \Longrightarrow DE=CD \tan(\angle DCE)=DB \tan(\angle DCE) \sin (\angle ABC-\theta). $$
Equating DE from each alternate path, $$ \tan(\angle DCE)\sin (\angle ABC-\theta)= \tan (\angle DAE) \sin \theta \qquad \Longrightarrow \\ \cot \theta= \cot (\angle ABC) + \frac{\tan (\angle DAE)}{\sin (\angle ABC)\tan (\angle DCE) } ~~. $$
Your other unknown angle follows then from $$ \tan (\angle DBE)=DE/DB= \tan (\angle DAE) \sin\theta = \tan (\angle DAE) \frac{1} {\sqrt{1+\cot^2\theta}} ~. $$