Maybe something quick (to set up, no guarantees of actual speed) and dirty.
Whatever your method was, it probably boiled down to a vector function
\begin{align}
\begin{bmatrix} x,\, y,\, z\end{bmatrix} = F\Big([s,\, t,\, u] \,\, \Big| \,\, V_{ijk} \, : i,j,k = 1..N\Big)
\end{align} where $V_{ijk} \in \mathbb{R}^3$are the vertices of the three dimensional lattice.
Basically, you currently have the new coordinates $[x,\, y,\, z]$ of a point, and you want to find the original ones $[s,\, t,\, u]$. I hope you still have the coordinates of the lattice vertices $V_{ijk}$. To retrieve the initial coordinates $[s,\, t,\, u]$ you have to solve for $[s,\, t,\, u]$ the system of equations
\begin{align}
\begin{bmatrix} x,\, y,\, z\end{bmatrix} = F\Big([s,\, t,\, u] \,\, \Big| \,\, V_{ijk} \, : i,j,k = 1..N\Big)
\end{align} where $[x,\, y,\, z]$ is given and the points $V_{ijk}$ are given
For simplicity, I am going to write
$$\begin{bmatrix} x,\, y,\, z\end{bmatrix} = F\big([s,\, t,\, u] \,\, \big| \,\, V \,\big)$$
To solve it, use a type of Newton's method.
Set a small enough number $h$.
Set your initial $s_0=x, t_0=y, u_0=z$ and set up your error equal to $err = 1$ (or whatever).
While $err \geq h^2$ execute the following iteration:
Assume you have a current approximation $[s_n,\, t_n,\, u_n]$;
Calculate the following three three-vector (as rows)
\begin{align}
&\Delta_s F_n = \frac{1}{h} \Big( \, F\big([s_n+h,\, t_n,\, u_n] \,\, \big| \,\, V \,\big) - F\big([s_n,\, t_n,\, u_n] \,\, \Big| \,\, V \,\big) \, \Big)\\
&\Delta_t F_n = \frac{1}{h} \Big( \, F\big([s_n,\, t_n+h,\, u_n] \,\, \big| \,\, V \,\big) - F\big([s_n,\, t_n,\, u_n] \,\, \Big| \,\, V \,\big) \, \Big)\\
&\Delta_u F_n = \frac{1}{h} \Big( \, F\big([s_n,\, t_n,\, u_n+h] \,\, \big| \,\, V \,\big) - F\big([s_n,\, t_n,\, u_n] \,\, \Big| \,\, V \,\big) \, \Big)
\end{align}
Set the following three by three matrix
$$M_n = \begin{bmatrix} \Delta_s F_n \\ \Delta_t F_n \\
\Delta_u F_n\end{bmatrix}$$ (the vectors are the ones computed above but transposed and are therefore vector-columns);
Calculate the inverse matrix $M_n^{-1}$;
Calculate the update
$$[s_{n+1},\, t_{n+1},\, u_{n+1}] = [s_n,\, t_n,\, u_n] + \Big(\,[x,\, y,\, z] - F\big([s_n,\, t_n,\, u_n] \,\, \Big| \,\, V \,\big)\,\Big)\, M_n^{-1}$$
(all of this is written as vector matrix multiplication, where you are working with $1 \times 3$ row-vectors and multiplying such$1 \times 3$ row-vectors with the $3 \times 3$ matrix $M_n^{-1}$ on the right)
Calculate the error $$err = \Big\|\, [x,\, y,\, z] - F\big([s_{n+1},\, t_{n+1},\, u_{n+1}] \,\, \Big| \,\, V \,\big)\, \Big\|^2$$
Go back to the beginning of the iteration to check whether $err \geq h^2$
The result of this iteration is an approximation of the coordinates $[s,\, t,\, u]$. I hope this works.
