Edit: I add below a proof of the existence.

I only have a proof of the unicity (not the existence so far), but it might already be useful.

In fact let me prove the following fact, which is equivalent to the unicity part of your question:

> Let $x$ be a positive matrix. If $d$ is a diagonal matrix such that $x+d$ is positive and such that $x^{-1}$ and $(x+d)^{-1}$ have the same diagonals, then $d=0$.

Proof: Since $d$ is diagonal, the trace of $d\left(x^{-1} - (x+d)^{-1}\right)$ is zero. But one can write this expression as `\[dx^{-1/2}\left(1- (1+x^{-1/2}dx^{-1/2})^{-1}\right)x^{-1/2},\]`
so that taking the trace and denoting by $a=x^{-1/2}dx^{-1/2}$, we get $0= Tr(a(1-(1+a)^{-1}))$.

If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $a$, the condition $x+d$ positive becomes $1+\lambda_i >0$, and the last equality becomes $0 = \sum \lambda_i(1-1/(1-\lambda_i)) = \sum \lambda_i^2/(1+\lambda_i)$, which is possible only if the $\lambda_i$ are all zero, i.e. $a=0$, i.e. $d=0$.

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In fact the existence is very easy just for topological reasons. More precisely, let me denote by $E: A \in M_n\mapsto  (\delta_{i,j} A_{i,j})$ the conditional expectation on the algebra of diagonal matrices, and by $f$ the map from $D=\{x \in M_n(\mathbb R), x>0\textrm{ and }E(x)=1\}$ to $I=\{x \in M_n, x=x^*\textrm{ and }E(x)=0\}$ defined by $f(x) = x^{-1} - E(x^{-1})$. You are asking whether $f$ is a bijection. And the answer is yes. I already proved that $f$ is injective. To prove that it is surjective, it is enough to prove that it is continuous, open and proper (because this would imply that the image is an open and closed subset of $I$, and hence everything since $I$ is connected). The continuity is obvious. $f$ is even differentiable, and the differential is explicitely computable and easily seen to be invertible at every point, so that $f$ is indeed open. It remains to check that it is proper.

The proof I have is not completely obvious, maybe I am missing something. Let me only sketch it. Take a sequence $x_k \in D$ that escapes every compact subset of $D$. Since $\|x\|\leq n$ for all $x \in D$, we have that $u_k=\|x_k^{-1}\|\to \infty$ (I consider the operator norm). We want to prove that $\|f(x_k)\|\to \infty$. Assume for contradiction that this is not the case, and that $\|f(x_k)\|\leq C$ for all $k$.

Let $\xi_k$ be a sequence of unit vectors in $\mathbb R^n$ such that $x_k \xi_k = 1/u_k \xi_k$. Now the key observation: the assumption that $\|f(x_k)\|\leq C$ implies that, for all diagonal matrix $d$ with $1$ or $-1$ on the diagonal, the distance from $d \xi_k$ to the space $E_k$ spanned by the eigenvectors of $x_k^{-1}$ relative to the eigenvalues in an interval $[u_k-O(1),u_k]$ goes to zero. This is because $E(x)$ is the average of $dxd$ over all such diagonal matrices. This implies that there is a vector $\eta_k$ in the canonical basis of $\mathbb R^n$ at distance $o(1)$ from $E_k$ (if $\xi_k(i)$ is the $i$-th coordinate of $\xi_k$, then $\xi_k(i) \eta_k$ is the average of $d\xi_k$ over all diagonal $d$'s with $i$-th term on the diagonal $1$ and $+-1$ on the other entries). In particular, $\|x_k \eta_k\|\to 0$. But this is a contradiction with the fact that since $E(x_k)=1$, $\langle x_k \eta_k, \eta_k\rangle = 1$.