I can interpolate a hyperbolic paraboloid given 4 points in space ($\pmb{a_1}, \pmb{a_2}, \pmb{b_1}, \pmb{b_2}$) with the following parametric equation for a ruled surface:
$$ \pmb{r}(u, v) = (1-v)((1-u)\pmb{a_1}+u\pmb{a_2}) + v((1-u)\pmb{b_1}+ u\pmb{b_2}) $$
Can I obtain the equation of the surface in cartesian coordinates?
You could painstakingly eliminate the parameters $u$ and $v$ (or have a computer do it for you), but if you’re performing the calculations yourself I think it might involve less work to transform the canonical hyperbolic paraboloid (“hypar”) $X^2-Y^2=Z$, represented by the matrix $$Q = \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&0&-\frac12\\0&0&-\frac12&0\end{bmatrix}.$$ Using your parameterization, this hypar can be generated by the points $$\mathbf a_1' = (-1,0,1)^T \\ \mathbf a_2' = (0,-1,-1)^T \\ \mathbf b_1' = (0,1,-1)^T \\ \mathbf b_2' = (1,0,1)^T,$$ so if $M$ maps these points to the ones that define the desired hypar, its associated matrix is $M^{-T}QM^{-1}$. The transformation matrix $M$ is easily constructed from the point pairs, but we really need $$M^{-1} = \begin{bmatrix}\mathbf a_1' & \mathbf a_2' & \mathbf b_1' & \mathbf b_2' \\ 1&1&1&1\end{bmatrix} \begin{bmatrix}\mathbf a_1 & \mathbf a_2 & \mathbf b_1 & \mathbf b_2 \\ 1&1&1&1\end{bmatrix}^{-1} = A'A^{-1}.$$ We can precompute part of $M^{-T}QM^{-1}=A^{-T}{A'}^TQA'A^{-1}$ to simplify the eventual computation a bit: $$Q'={A'}^TQA' = \begin{bmatrix}0&0&0&-2\\0&0&2&0\\0&2&0&0\\-2&0&0&0\end{bmatrix}$$ so that the matrix of the hypar is given by the product $A^{-T}Q'A^{-1}$. Since we’re working in homogeneous coordinates, we can multiply this matrix by $1/2$ (or any other nonzero scalar) for convenience without changing the resulting quadric.
For example, let $$\mathbf a_1 = (-2,-1,2)^T \\ \mathbf a_2 = (-1,1,0)^T \\ \mathbf b_1 = (2,-2,1)^T \\ \mathbf b_2 = (3,0,2)^T.$$ We then have $$A^{-1} = \begin{bmatrix}-2&-1&2&3\\-1&1&-2&0\\2&0&1&2\\1&1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}-\frac5{27}&-\frac2{27}&\frac13&-\frac19 \\ -\frac1{27}&\frac5{27}&-\frac13&\frac79 \\ \frac2{27}&-\frac{10}{27}&-\frac13&\frac49\\ \frac4{27}&\frac7{27}&\frac13&-\frac19 \end{bmatrix}.$$ Multiplying this by $27$ to eliminate the fractions and using $\frac12Q'$ yields $$\begin{bmatrix}36&63&0&27\\63&-72&0&-135\\0&0&0&-243\\27&-135&-243&486\end{bmatrix}.$$ Pulling out another factor of $9$, this corresponds to the equation $$4x^2+14xy-8y^2+6x-30y-54z+54=0.$$ (Another factor of $2$ can be pulled out of this equation, if desired.) Plotting both the parametric patch and this hypar in Geogebra produces the following:
You could instead start from the canonical equation $XY=Z$, with associated matrix $$\begin{bmatrix}0&\frac12&0&0\\\frac12&0&0&0\\0&0&0&-\frac12\\0&0&-\frac12&0\end{bmatrix}$$ and go through a similar calculation after selecting a suitable set of points to generate this hypar. If you precompute $Q'$ as above, you should end up with a scalar multiple of the previous matrix.