Assuming all the $\alpha_j$ are nonzero, the matrices $X$ are Cauchy-like matrices, since you can rewrite them as $$ X_{ij} = \frac{\alpha_j^{-1}}{\alpha_j^{-1}-\bar{\alpha}_i} $$ so there are analogous formulas for their determinant and inverse. Note that $XA^{-1}$ is a Cauchy matrix, where $A = diag(\alpha_i)$, so these formulas follow directly from those for Cauchy matrices.

A variant of this equivalence that does not require the invertibility of $A$ and better exploits symmetry/Hermitianity is the following. Let $B = (I+A)(A-I)^{-1}$ (note that $\alpha_i\neq 1$, otherwise there would be a zero denominator in $X_{ii}$, so $A-I$ is invertible). Then, $$ B^*X+XB = -2(A- I)^{-*}E(A- I)^{-1} $$ expands to $$ ( I+A^*)X(A- I) + (A^*- I)X( I + A) = -2E $$ which reduces to $$ 2A^*XA - 2X = -2E, $$ which is the Stein equation that you used to define $X$.

So $X$ solves a Lyapunov equation with a diagonal $B$, and hence one can write the more symmetric formula

$$X_{ij} = \frac{-2(\alpha_i-1)^{-*}(\alpha_j-1)^{-1}}{\bar{\beta}_i + \beta_j},$$

where $\beta_i = \frac{\alpha_i + 1}{\alpha_i-1}$ are the diagonal entries of $B$. Note that $\Re\beta_i < 0$ iff $|\alpha_i| < 1$.

Alternatively, the Hermitian matrix $(A^*-I)^{-1}X(A-I)^{-1}$ is a Cauchy matrix wrt the two sequences $\bar{\beta}_i$ and $-\beta_j$.

This trick to convert between Lyapunov and Stein equations is classical (bilinear transform, or Cayley transform).

Assuming all the $\alpha_j$ are nonzero, the matrices $X$ are Cauchy-like matrices, since you can rewrite them as $$ X_{ij} = \frac{\alpha_j^{-1}}{\alpha_j^{-1}-\bar{\alpha}_i} $$ so there are analogous formulas for their determinant and inverse. In particular, $XA^{-1}$ is a Cauchy matrix, where $A = diag(\alpha_i)$, so these formulas follow directly from those for Cauchy matrices.

A variant of this equivalence that does not require the invertibility of $A$ and better exploits symmetry/Hermitianity is the following. Let $B = (I+A)(A-I)^{-1}$ (note that $\alpha_i\neq 1$, otherwise there would be a zero denominator in $X_{ii}$, so $A-I$ is invertible). Then, $$ B^*X+XB = -2(A- I)^{-*}E(A- I)^{-1} $$ expands to $$ ( I+A^*)X(A- I) + (A^*- I)X( I + A) = -2E $$ which reduces to $$ 2A^*XA - 2X = -2E, $$ which is the Stein equation that you used to define $X$.

So $X$ solves a Lyapunov equation with a diagonal $B$, and hence one can write the more symmetric formula

$$X_{ij} = \frac{-2(\alpha_i-1)^{-*}(\alpha_j-1)^{-1}}{\bar{\beta}_i + \beta_j},$$

where $\beta_i = \frac{\alpha_i + 1}{\alpha_i-1}$ are the diagonal entries of $B$. Note that $\Re\beta_i < 0$ iff $|\alpha_i| < 1$.

Alternatively, the Hermitian matrix $(A^*-I)^{-1}X(A-I)^{-1}$ is a Cauchy matrix wrt the two sequences $\bar{\beta}_i$ and $-\beta_j$.

This trick to convert between Lyapunov and Stein equations is classical (bilinear transform, or Cayley transform).