It happens that the $3$-dimensional coordinates are far nicer than their $2$-dimensional counterparts. For another example of this kind of thing, consider the equilateral triangle: in the plane, sample coordinates are: $(1,0)$, $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$, $\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$; not too bad, but space offers us the ultra-simple $(1,0,0)$, $(0,1,0)$, $(0,0,1)$.
As for a tetartoid pentagon $ABCDE$, we have (slightly modified) Ed Pegg's parameterization of these $3$d coordinates:
$$A = (a, b, c) \qquad B = (−a, −b, c) \qquad C = \frac{cm}{p}(-1,-1,1) \\ D = (−c, −a, b) \qquad E = \frac{cm}{q}(-1,1,1) \tag{1}$$
where $0 \leq a \leq b \leq c$ and
$$m := b c - a^2 \geq 0 \qquad p := m + ( b - a ) ( c - b ) \geq 0 \qquad q := m + ( a + b ) ( c-b ) \geq 0$$
(As you note, the $a$, $b$, $c$ here are not used in the same way as in the "rhyme scheme" $abbcc$ for the congruence of edge-lengths in the pentagon. They're simply independent parameters.) The coordinates above are reasonably uncomplicated; converting them to $2$d isn't too difficult a process, but yields messier expressions:
$$
A = (-r,0) \qquad B = (r, 0) \tag{2}\\
C = \frac{c}{pr} \left(\; m ( a + b ), s ( b - a )\;\right) \quad
D = \frac{1}{r}\left(\; a ( b + c ), s \;\right) \quad
E = \frac{c}{qr}\left(\; -m ( b - a ), s ( a + b ) \;\right)
$$
where $r := \sqrt{a^2+b^2}$ and $s := \sqrt{( a^2 - b c )^2 + ( a^2 + b^2 ) ( b - c )^2}$.
We can simplify the coordinates in $(2)$ slightly by magnifying the pentagon by a factor of $\sqrt{a^2+b^2}$ (that is, by multiplying everything by $r$, which eliminate our need for "$r$" altogether):
$$
A = (-(a^2+b^2),0) \qquad B = (a^2+b^2, 0) \tag{3}\\[8pt]
C = \frac{c}{p} \left(\; m ( a + b ), s ( b - a )\;\right) \quad
D = \left(\; a ( b + c ), s \;\right) \quad
E = \frac{c}{q}\left(\; -m ( b - a ), s ( a + b ) \;\right)
$$
It's possible that further simplification is possible ---perhaps in terms of parameters other than $a$, $b$, $c$--- but $(3)$ is the best I have at the moment.
