The linked question also asks for the special case that the off-diagonal elements of $\Sigma^{-1}$ have a rank-one structure, $(\Sigma^{-1})_{ij}=u_i v_j$ for $i\neq j$. In that case the covariance matrix can be reconstructed as follows.

Denote $\Sigma^{-1}=A+uv^\top$, with $u,v$ known rank-$N$ vectors and $A={\rm diag}\,(d_1,d_2,\ldots d_N)$ an unknown diagonal matrix. According to the Sherman–Morrison formula, the covariance matrix $\Sigma$ then has diagonal elements $$\Sigma_{ii}= x_i - {x_i^2 u_iv_i \over 1 +\sum_{n=1}^N u_nv_n x_n },\;\;i=1,2,\ldots N,$$ with $\;\;x_n=1/d_n$. These are $N$ equations where the $N$ unknowns $x_n$ appear quadratically.

This method solution seems efficient, but I do not have a proof that a unique solution will be obtained.

In the linked question the OP also asks for the special case that the off-diagonal elements of $\Sigma^{-1}$ have a rank-one structure, $(\Sigma^{-1})_{ij}=u_i v_j$ for $i\neq j$. In that case the covariance matrix can be reconstructed as follows.

Denote $\Sigma^{-1}=A+uv^\top$, with $u,v$ known rank-$N$ vectors and $A={\rm diag}\,(d_1,d_2,\ldots d_N)$ an unknown diagonal matrix. According to the Sherman–Morrison formula, the covariance matrix $\Sigma$ then has diagonal elements $$\Sigma_{ii}= x_i - {x_i^2 u_iv_i \over 1 +\sum_{n=1}^N u_nv_n x_n },\;\;i=1,2,\ldots N,$$ with $\;\;x_n=1/d_n$. These are $N$ equations where the $N$ unknowns $x_n$ appear quadratically.

This method of solution is efficient, but I do not have a proof that the solution with a positive definite $\Sigma$ is unique.