Since this question remained without answers even after a bounty, I thought it might be time to ask it here.
For which pyramid can you compute the volume from simple cut-and-glue processes? The Dehn invariant naturally gives the answer, but I failed to turn this in an algorithm. Here are the pyramids, I know of, whose volume is computable by elementary operations:
take a cube and divide into six pyramids from its center (or three pyramids given the 120° symmetry along a diagonal). These pyramids can be subdivided and glued into further pyramids, but there are not so many possibilities.
take a trigonal trapezohedron whose faces are all rhombic. This is not a right prism, but oblique prisms can also be cut-and-glued into a right prism (and hence their volume can be computed). Because of his symmetry group you can cut it into three pyramidal pieces (oblique pyramids). Since the small angle of the rhombus can be $\in ]0, \pi/2[$ (at $\pi/2$ it's just a cube), this gives an infinite family of pyramids. (These oblique pyramids have a symmetry: you can cut them further in half.) EDIT: Note that when the acute angle is $\in ]\pi/3, \pi/2[$ there are actually two ways to make a trigonal trapezohedron out of this rhombus (depending on whether the angle at the polar vertices is acute or obtuse).
But the that's all I could find. Are there any other? and if so, what is the cut-and-glue process?
Some background:
The volume of pyramids is discussed here in Proposition 3 to 5 of Euclid's book
The fact there was no "simple" proof irritated Gauß and later Hilbert. It was purportedly the inspiration for his 3rd Problem, which was solved by Dehn (using his invariant)
There is an earlier (and largely ignored) solution of Hilbert's 3rd problem by a certain Birkenmayer, see this paper.
In two dimensions the Wallace–Bolyai–Gerwien theorem answers the question completely. Beside the Dehn invariant (which shows this is not the case in three dimensions), the Banach-Tarski paradox is also a reminder that volume in three dimensions are tricky.
[EDIT: To clarify simple cute-and-glue (as asked in the comments). One starts with the class $(C)$ of solids (at the beginning it only includes rectangular prisms). You can add another polyhedron $P$ to $(C)$ if:
a number $k \neq 0$ of copies of $P$ can be obtained by cutting solids $\{S_i\}_{i=1,\ldots,s}$ from $(C)$ [possibly many copies of the same solid] into finitely many polyhedral pieces and reassembling these pieces together into the $k$ copies $P$ and a possibly empty collection of solids $\{T_i\}$ with $T_i$ belonging to $(C)$. In that case, $\mathrm{vol}(P) = \tfrac{1}{k} \big( \sum \mathrm{vol}(S_i) - \sum \mathrm{vol}(T_i) \big)$, and so $P$ can be added to $(C)$.
Reassembling means apply isometries of $\mathbb{R}^3$ (rotations, reflections and translations); two solids are congruent if one is the image of the other under such transformations and a copy is also the image of a solid under such transformations. This definition is closely related to scissors-congruence.
