Usually, we have a matrix $A$ and want to calculate the $LU$ (or sometimes Cholesky, depending on $A$'s properties) decomposition. This is often the hard part. Now, if we have the $LU$ decomposition of $A$, it is easy to calculate $A^{-1}$ (triangular matrices are easy to invert - FW backward substitution is stable and can be parallelized).
Is the other way also possible? I.e. given $A$ and $A^{-1}$, can we calculate the LU decomposition (or Cholesky) of $A$ stable, and easy to parallelize, in a similar way as inverting triangular matrices?
I thought about it for a little while, I can't see it. I can't prove it is impossible, but it is not apparent by elementary arguments. For simplicity of discussion, suppose $A$ admits a $LU$ decomposition.
Consider, $A=LU$ then $A^{-1} = U^{-1}L^{-1}$ so, if I take products of the given $A$ and $A^{-1}$ I can not see how to isolate $L$ and $U$ since $AA^{-1} = LUU^{-1}L^{-1}=I$ yields no insight.
In principle, $A^{-1}$ is the product of elementary matrices. The matrices $E_1, \dots E_k$ accomplish the so-called forward pass which gives $ref(A) = E_k \cdots E_1A$ where $ref(A)$ has zeros below the pivot positions. Given $A^{-1}$ exists, each column is a pivot so we can assume that $ref(A) = I + N$ where $N$ is a strictly upper triangular matrix and $I$ is the identity. Next, the Gaussian elimination proceeds to the backwards pass which clears out the nontrivial entries in $N$. We can view the backwards pass as multiplication by elementary matrices $E_{k+1}, \dots ,E_{r}$. We find: $$ E_{r} \cdots E_{k+1}E_{k} \cdots E_{1}A = E_{r} \cdots E_{k+1}ref(A)=I $$ since $rref(A)=I$ given $A^{-1}$ exists. The inverse is unique, so we have $$ A^{-1} = E_{r} \cdots E_{k+1}E_{k} \cdots E_{1} $$ But, how do we pull-apart the $L$ and $U$ from this list? Given my notation, they are: $$ U = ref(A) = E_{k} \cdots E_{1}A \qquad \text{and} \qquad L = E_{k+1}^{-1} \cdots E_{r}^{-1} $$ The problem is that $A^{-1}$ has both the forward and backwards pass implicit within it's structure. I can't see how to separate them without some other insight.
Fortunately, we do have a method to factor a matrix $A$ into the $LU$ (or sometimes $PLU$) so... while I like your question, I'm not sure how to go further with it.