I have considered two different paths to this problem, one requires the diagonalization of $A$ (so may be not so useful depending on $A$) and the other uses the Kronecker product properties. One other note at the end gives an equivalent skew form of the equation. The Kronecker seems to have the most usefulness for finding the desired trace. The rest is maybe something of interest worth looking at.
First, the Kronecker. Using
$\otimes$ to denote the Kronecker product, the vectorization of the formula gives
$A\Theta + \Theta A^\top = 2D \iff \left( I \otimes A + A \otimes I\right) \operatorname{vec}(\Theta) = \operatorname{vec}(2D)$
That together with a formula for the trace of a product of four matrices (the first line of the following), we get
\begin{align} \operatorname{Trace}(A^\top D^{-1} A \Theta) &= \left( \operatorname{vec}(D^{-\top})\right)^\top\left(A^\top \otimes A\right)\operatorname{vec}(\Theta) \\ &=\left( \operatorname{vec}(D^{-\top})\right)^\top\left(A^\top \otimes A\right)\left(I\otimes A + A \otimes I\right)^{-1}\operatorname{vec}(2D) \end{align} This may be useful in that it is a formula using only the knowns, though it is of higher dimension than the original.
The next path attempts to keep in terms of the original dimensions, though requires the diagonalization of A. Let
$D = EE^\top$ since it is symmetric positive definite. This is found with use of the Cholesky factorization. Then we have: \begin{align} A\Theta + \Theta A^\top &= 2EE^\top \\ \end{align}
Let
$A$ be diagonalizable with
$A = Q \Lambda Q^{-1}$ for some diagonal
$\Lambda$ \begin{align} Q \Lambda Q^{-1}\Theta + \Theta \left(Q \Lambda Q^{-1}\right)^\top &= 2EE^\top \\ Q \Lambda Q^{-1}\Theta + \Theta Q^{-\top} \Lambda^\top Q^\top &= 2EE^\top \\ \end{align}
Multiply the entire equation: $Q^{-1}\left( \dots \right) Q^{-\top}$: \begin{align} \Lambda Q^{-1}\Theta Q^{-\top} + Q^{-1}\Theta Q^{-\top} \Lambda^\top &= 2Q^{-1}EE^\top Q^{-\top} \\ \end{align} Use substitution $X = Q^{-1}\Theta Q^{-\top}$ ( noting that $\Lambda^\top = \Lambda$): \begin{align} \Lambda X + X \Lambda &= 2Q^{-1}EE^\top Q^{-\top} \\ \end{align}
Solving for $X$ here requires an element by element inspection. Let $C = 2Q^{-1}EE^\top Q^{-\top} $. Then element $i,j$ has equation: \begin{align} \left[ \Lambda X + X \Lambda \right]_{ij} &= C_{ij} \\ \Lambda_{ii}X_{ij} + X_{ij}\Lambda{jj} &= C_{ij} \\ \end{align} This may be seen since left multiplication by a diagonal scales the rows and right multiplication scales the columns. So for each element of $X$ we have:
$X_{ij} = \frac{C_{ij}}{\Lambda_{ii} + \Lambda_{jj}}$
Or in matrix form: $\left[ X_{ij}\right] = \left[ \frac{C_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right] = X$ In terms of $\Theta$: $\Theta = Q \left[ \frac{C_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right]Q^\top = Q \left[ \frac{\left[2Q^{-1}EE^\top Q^{-\top}\right]_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right]Q^\top$
Let us look now at $A \Theta A^\top$, using $A = Q \Lambda Q^{-1}$ $A \Theta A^\top = Q \Lambda Q^{-1} \left( Q \left[ \frac{\left[2Q^{-1}EE^\top Q^{-\top}\right]_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right]Q^\top\right)\left(Q \Lambda Q^{-1}\right)^\top$
$A \Theta A^\top = Q \Lambda \left[ \frac{\left[2Q^{-1}EE^\top Q^{-\top}\right]_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right] \Lambda Q^\top$
Now for $\operatorname{Trace}(D^{-1}A \Theta A^\top) = \operatorname{Trace}(E^{-\top}E^{-1}A \Theta A^\top) = \operatorname{Trace}(E^{-1}A \Theta A^\top E^{-\top}) $ (using $D=EE^\top$ and the cyclic property of trace):
$\operatorname{Trace}(E^{-1}A \Theta A^\top E^{-\top}) = \operatorname{Trace}(E^{-1}Q \Lambda \left[ \frac{\left[2Q^{-1}EE^\top Q^{-\top}\right]_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right] \Lambda Q^\top E^{-\top}) $
The $\Lambda$ on the left gives row scaling and on the right gives column scaling, for $\operatorname{Trace}(D^{-1}A \Theta A^\top) = \operatorname{Trace}(E^{-1}Q \left[ \frac{\Lambda_{ii}\Lambda_{jj}\left[2Q^{-1}EE^\top Q^{-\top}\right]_{ij}}{\Lambda_{ii} + \Lambda_{jj}}\right] Q^\top E^{-\top}) $
From here ...? Note that $D=EE^\top$ is congruence transformed, scaled (element $ij$ by $\frac{2\Lambda_{ii}\Lambda{jj}}{\Lambda_{ii} + \Lambda_{jj}}$)then inverse congruence transformed to get the trace equivalent of $D^{-1}A\Theta A^\top$. Not sure exactly how useful that is though.
One other possibly interesting note, since you know that
$\Theta$ is symmetric (your statement or also may be seen from above) then
$\Theta = BB^\top$ for some
$B$. This gives \begin{align} ABB^\top + BB^\top A^\top &= 2D \\ \end{align} Using
$Y=ABB^\top$ and
$Y^\top = BB^\top A^\top$ we have
$Y+Y^\top = 2D$ and
$Y^\top+Y = 2D^\top = 2D$ (since
$D$ symmetric is given).
$Y$ then must be of the form of a skew symmetric plus
$D$:
$ Y = S + D$ with
$S^\top = -S$. This gives two other forms for
$\Theta$ as
$\Theta = BB^\top = A^{-1}Y = A^{-1}\left( S + D\right)$ and
$\Theta = BB^\top = Y^\top A^{-\top} = \left( -S + D\right)A^{-\top}$ Plugged into the original equation checks that it works: \begin{align} A\Theta + \Theta A^\top &= 2D \\ AA^{-1}\left( S + D\right) + \left( -S + D\right)A^{-\top}A^\top &= 2D \\ \left( S + D\right) + \left( -S + D\right) &= 2D \\ 2D &= 2D \end{align} Looking at the two forms
$\Theta = \Theta$ \begin{align} A^{-1}\left( S + D\right) = \left( -S + D\right)A^{-\top} \\ \left( S + D\right) = A\left( -S + D\right)A^{-\top} \\ \left( S + D\right)A^\top = A\left( -S + D\right) \\ \end{align} and a formula for the skew matrix
$S$ is realized: \begin{align} SA^\top + AS &= AD - DA^\top \\ \end{align}