I'm trying to construct a parallelepiped where each face has diagonal lengths $d_1$ and $d_2$. I start by constructing a parallelogram in the $x,y$ -plane that has the correct diagonals. I put that it has side vectors $(a, 0, 0)$ and $(b1, b2, 0)$ where
$$ \begin{align} b_1 &= \frac{d_2^2-d_1^2}{4a} \\ b_2 &= \frac{1}{4a}\sqrt{(d_2^2+d_1^2)a^2-16a^4-(d_2^2-d_1^2)^2} \end{align}$$
and $a$ is a free parameter (restricted to an interval). This seems to work according to this Desmos graph.
Then let the third side vector be $(c_1, c_2, c_3)$. From the diagonal lengths we get the following equations
$$\begin{align} (c_1+a)^2 + c_2^2 + c_3^2 &= d_2^2 \\ (c_1-a)^2 + c_2^2 + c_3^2 &= d_1^2 \\ (c_1+a)^2 + (c_2+b_2)^2 + c_3^2 &= d_2^2 \\ (c_1-a)^2 + (c_2-b_2)^2 + c_3^2 &= d_1^2 \end{align}$$
And solving them (subtract the top two to get $c1$ and then the bottom two to get $c_2$) I get
$$\begin{align} c_1 &= b_1\\ c_2 &= \frac{d_2^2-d_1^2 - 4b_1^2}{4b_2} \\ c_3 &= \sqrt{d_1^2-(c_1-a)^2-c_2^2} \end{align}$$
But then we also need to satisfy $(c_1+a)^2 + (c_2+b_2)^2 + c_3^2 = d_2^2$ with this $c_3$ and get the enormous equation for $a$
$$ \frac{d_2^2-d_1^2}{2} - \frac{(d_2^2-d_1^2)^2}{8a^2} + \frac{1}{16a^2}(8(d_2^2+d_1^2)a^2-16a^4-(d_2^2-d_1^2)^2) = 0 $$ That is a quadratic in $a^2$ and in the end we get only one (or possibly $4$) working $a$. Is this correct?
(At first I didn't realize that we still need to check for satisfying the bottom equations and not any $a$ will actually work, but now as to use the only correct working $a$ it seems to go through. But I'm too lazy to solve the equation and I'm posting this anyhow.)
It is easy to show that all faces of the parallelepiped are congruent rhombi.
Consider a vertex of the parallelepiped where the equal diagonals meet (there are two such vertices). Let the edges be $a,b,c$ with $\gamma,\alpha,\beta$ being the respective angles between the pairs of the edges at the vertex.
Then we have the following system of equations: $$ a^2+b^2+2ab\cos\gamma=d_1^2,\quad a^2+b^2-2ab\cos\gamma=d_2^2,\\ b^2+c^2+2bc\cos\alpha=d_1^2,\quad b^2+c^2-2bc\cos\alpha=d_2^2,\\ c^2+a^2+2ca\cos\beta=d_1^2,\quad c^2+a^2-2ca\cos\beta=d_2^2.\\ $$
This implies $$a^2+b^2=b^2+c^2=c^2+a^2=\frac{d_1^2+d_2^2}2\implies a=b=c =\frac{\sqrt{d_1^2+d_2^2}}2$$ and $$ \alpha=\beta=\gamma=2\arctan\frac{d_2}{d_1}. %\tan\frac\alpha2=\tan\frac\beta2=\tan\frac\gamma2=\frac{d_2}{d_1}. $$
From the last equation it follows that the parallelepiped exists only if $\displaystyle\frac{d_2}{d_1}<\sqrt3$.