According to the comments the only ugly part was the devectorization.
Consider the 3x3 matrix $${\bf X}=\left[\begin{array}{ccc}
1&2&3\\4&5&6\\7&8&9\end{array}\right]$$
It's lexical vectorization is:
$$\text{vec}({\bf X}) = \left[\begin{array}{ccccccccc}1&4&7&2&5&8&3&6&9\end{array}\right]^T$$
We consider that kind of vectorization here.
The Matlab / Octave expression:
kron([1,0,0]',eye(3))'*kron(R(:),[1,0,0])+
kron([0,1,0]',eye(3))'*kron(R(:),[0,1,0])+
kron([0,0,1]',eye(3))'*kron(R(:),[0,0,1])-R
evaluates to the zero matrix for several random matrices R, each new term "selecting" a new row.
Then a guess would be that
$${\bf R}=\sum_k({\bf v_k} \otimes {\bf I})^T(\text{vec}({\bf R})\otimes {\bf v_k}^T)$$
With $\bf v_k$ being the Dirac or selector vector:
$$({\bf v_k})_i = \cases{0, k \neq i\\1, k=i}$$
Feel free to try and simplify it if you want.
Schur product theorem states that Hadamard product of two positive semidefinite matrices is positive semidefinite.
$B$ is negative definite $\implies -B$ is positive definite.
Since $$A \circ B = -(A \circ (-B)),$$ and $A \circ (-B)$ is positive semidefinite by Schur product theorem.
We conclude that $A \circ B$ is negative semidefinite.
Best Answer
There won't be such a formula. In fact, if $A$ and $B$ are invertible, we can't guarantee that $A\circ B$ will even have an inverse. In particular, consider $$ A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{0&1\\1&0} $$ we note that $A=A^{-1}$ and $B = B^{-1}$, but $A\circ B = 0$.
It is often, however, useful to consider the Hadamard product as a submatrix of the Kronecker product. For the Kronecker product, we have $$ (A \otimes B)^{-1} = A^{-1} \otimes B^{-1} $$