That such a decomposition doesn't exist can be shown by looking at the interior solid angles at the vertices.
The cube has eight interior solid angles of $\pi/2$ each.
Each of the six tetrahedra has one interior solid angle of $\pi/2$ and three that can be computed as shown here:
$$\Omega=\phi_{ab}+\phi_{bc}+\phi_{ac}-\pi=\pi/2 + \arccos\frac{1}{\sqrt{3}}+ \arccos\frac{1}{\sqrt{3}}-\pi=2\arccos\frac{1}{\sqrt{3}}-\pi/2\;.$$
Thus, the sum of the interior solid angles of the cube is $4\pi$, and the sum of the interior solid angles of the tetrahedra is
$$6\left(\pi/2+3\left(2\arccos\frac{1}{\sqrt{3}}-\pi/2\right)\right)=6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)\;.$$
This is not equal to $4\pi$, since $\arccos\frac{1}{\sqrt{3}}$ is not a rational multiple of $\pi$. (Here's an elementary proof of that fact.) In fact it is almost $5\pi$.
[Edit: I just realized in the shower that all the following analysis of potential inner vertices is actually not necessary. No matter what happens at the inner vertices, we have to fill the eight vertices of the cube with vertices of the tetrahedra, and the six vertices with interior solid angle $\pi/2$ are not enough to do it, so we have to use at least one of the others, but then we can't make the sum come out to $4\pi$, a rational multiple of $\pi$.]
This establishes that you can't fill the cube with the six tetrahedra by letting all their vertices coincide with the cube's vertices. It seems geometrically obvious that you can't do it with any of the vertices inside the cube, either, but this, too, can be proved rigorously using the solid angles. If there were a vertex inside the cube, the entire solid angle of $4\pi$ around that inner vertex would have to be filled. The candidates for filling it are a face of a tetrahedron, which subtends a solid angle of $2\pi$, an edge of a tetrahedron, which subtends a solid angle of twice the dihedral angle of the intersecting planes, i.e. either $\pi$ or $2\arccos\frac{1}{\sqrt{3}}$, or an interior solid angle of a vertex of a tetrahedron. Thus, adding up all the solid angles at the eight outer vertices and at $v$ inner vertices, the following equation would have to have solutions with non-negative integer values of $j$, $k$, $l$, $m$, $n$ and $v$:
$$6\left(6\arccos\frac{1}{\sqrt{3}}-\pi\right)+j(2\pi)+k(\pi)+l\left(2\arccos\frac{1}{\sqrt{3}}\right)+m(\pi/2)+n\left(\arccos\frac{1}{\sqrt{3}}-\pi/2\right)$$
$$=4\pi+v(4\pi)\;.$$
This is impossible, since the coefficient in front of $\arccos\frac{1}{\sqrt{3}}$ is non-zero, and thus the equation would imply that this is a rational multiple of $\pi$, which it isn't (see above).
P.S.: I just realized I forgot to mention an essential part of the proof:
$$\cos 2\phi = 2 \cos^2 \phi - 1\;,$$
$$\arccos x = \frac{1}{2}\arccos(2x^2-1)\;,$$
$$\arccos \frac{1}{\sqrt{3}}=\frac{1}{2}\arccos(-\frac{1}{3})\;,$$
so $\arccos \frac{1}{\sqrt{3}}$ is rational iff $\arccos(-\frac{1}{3})$ is rational; then we can apply the theorem I linked to above.
So you came up with a formula which is (with very little rearrangement)
$$V = \frac{(xyz)^2}{2}+2 xyz+\frac{(xyz)^3}{24}+\frac83,$$
and you know from the very beginning that
$$xyz = 2.$$
Having those two pieces of information, you are a very short step away from
the solution. You should be able to use a simple substitution to find the
numeric value of the first formula.
The second formula tells you a substitution you can use.
I don't know exactly how you got the first formula; a comparison to a more
general formula for $xyz = c,$ where $c$ is an arbitrary positive constant,
makes me think you already used the specific fact that $xyz = 2.$
(Which is perfectly OK, because you were given that fact in the problem statement.)
Since the resulting volume for $xyz = 2$ agrees with the general solution,
however, I don't see any reason to suspect that you made any errors.
You can also look at the solutions and links mentioned in
Show that product of x, y, and z intercents of tangent plane to surface xyz=1 is a constant.
Your question is almost a duplicate of that one (up to a constant factor),
except that you had gotten much closer to a solution
at the point where you asked your question.
Best Answer
The volume of a tetrahedron satisfies
where $a$, $b$, $c$ are lengths of edges coinciding at a vertex, and $\alpha$, $\beta$, $\gamma$ are the angles between those edges ($\alpha$ between $b$ and $c$, etc).
For the tetrahedron in question, we see that $a$, $b$, $c$ and $\alpha$, $\beta$, $\gamma$ are also elements of a triangle. Since $\alpha+\beta+\gamma = 180^\circ$, the trig factor of $(1)$ reduces, and we have $$36 V^2 = a^2 b^2 c^2 \cdot 4 \cos \alpha \cos \beta \cos \gamma \quad\to\quad 9V^2 = a^2b^2c^2\cos\alpha\cos\beta\cos\gamma \tag{2}$$ which, by the Law of Cosines, we can write as
Another way to get at this result is with the Pseudo-Heron Formula for volume. (See my note, "Heron-like Hedronometric Results for Tetrahedral Volume" (PDF).) If $W$, $X$, $Y$, $Z$ are the face-areas of a tetrahedron, and $H$, $J$, $K$ are the pseudoface-areas (see below) then
In an equihedral tetrahedron ($W=X=Y=Z$), this reduces to
A pseudoface of a tetrahedron is the quadrilateral shadow of the figure in a plane parallel to two opposite edges. If edges $a$, $b$, $c$ meet at a vertex, and have respective opposite edges $d$, $e$, $f$, then our $H$, $J$, $K$ are related to respective edge-pairs $(a,d)$, $(b,e)$, $(c,f)$, and we have, for instance, $$16 H^2 = 4 a^2 d^2 - \left(\; b^2 - c^2 + e^2 - f^2 \;\right)^2 \tag{6}$$ In the tetrahedron in question, $a=d$, $b=e$, $c=f$, so that $$\begin{align} 4 H^2 = a^4 - \left( b^2 - c^2 \right)^2 &= \left(\phantom{-}a^2 - b^2 + c^2 \right)\left(\phantom{-}a^2 + b^2 - c^2 \right)\\ 4 J^2 \;\;\phantom{= a^4 - \left( b^2 - c^2 \right)^2} &= \left(\phantom{-}a^2+b^2-c^2\right)\left(-a^2+b^2+c^2\right) \\ 4 K^2 \;\phantom{= a^4 - \left( b^2 - c^2 \right)^2} &= \left( -a^2+b^2+c^2\right)\left(\phantom{-}a^2-b^2+c^2\right) \end{align}\tag{7}$$ and then $(3)$ follows from $(5)$.
Edited to add some hedronometric context to @Calum's answer ...
The rectangular faces of the tetrahedron's bounding cuboid are exactly the figure's pseudofaces. Indeed, we can deduce that the pseudofaces must be rectangles: each corresponding pseudofacial projection of a tetrahedron with three pairs of opposite congruent edges is necessarily a quadrilateral with two pairs of opposite congruent edges (hence, a parallelogram) and a pair of congruent diagonals (hence, a rectangle).
Specifically, the rectangular pseudoface $H$ (the projection into a plane parallel to the $a$ edges) has congruent diagonals $a$, and edges $b^\prime := \sqrt{b^2-h^2}$ and $c^\prime := \sqrt{c^2-h^2}$, where $h$ is distance between planes containing the $a$ edges (that is, $h$ is the corresponding "height" of the cuboid, what I call a pseudoaltitude of the tetrahedron). Since $a^2 = (b^\prime)^2 + (c^\prime)^2$ in the rectangle, we have that $h^2=\left(-a^2+b^2+c^2\right)/2$, whence $b^\prime = \sqrt{\left(a^2-b^2+c^2\right)/2\;}$ and $c^\prime=\sqrt{\left(a^2+b^2-c^2\right)/2\;}$. As $H = b^\prime c^\prime$, we reconfirm equation $(7)$.
As shown in @Calum's cuboid figure, each face of the given tetrahedron is the "hypotenuse-face" of a right-corner tetrahedron whose "leg-faces" are the three half-pseudofaces. (I hadn't really noticed this before!) By de Gua's Theorem , $$\left(\frac12H\right)^2+\left(\frac12J\right)^2+\left(\frac12K\right)^2 = W^2 \quad\to\quad H^2 + J^2 + K^2 = 4 W^2$$ This is consistent with the Sum of Squares identity that holds for any tetrahedron: $$H^2 + J^2 + K^2 = W^2 + X^2 + Y^2 + Z^2$$
@Calum expresses the tetrahedron's volume in terms of pseudoaltitudes, which I tend to label $h$, $j$, $k$, instead of $x$, $y$, $z$; my $(5)$ expresses the volume in terms of pseudoface areas $H$, $J$, $K$. As it happens, any tetrahedron's volume is given by each pseudoface-pseudoaltitude pair: $$3 V = Hh = Jj = Kk$$ This gives us a bridging relation between the two formulas: $$27 V^3 = H J K \cdot h j k$$