For $k \in \mathbb{N}$ the covariance of $x_k$ is given by $\mathbb{E}\left[ x_k x_k^\top \right] - \mathbb{E}\left[ x_k \right] \mathbb{E}\left[ x_k \right]^\top$ where
$$\mathbb{E}\left[ x_k x_k^\top \right] = A^{k} \mathbb{E}\left[ x_0 x_0^\top \right] \left( A^k \right)^\top + \sum_{i=0}^k A^i B R_w B^\top \left( A^\top \right)^i$$
and
$$\mathbb{E}\left[ x_k \right] \mathbb{E}\left[ x_k \right]^\top = A^k \mathbb{E}\left[ x_0 \right] \mathbb{E}\left[ x_0 \right]^\top \left( A^\top \right)^k$$
so the covariance is
$$A^{k} \left( \mathbb{E}\left[ x_0 x_0^\top \right] - \mathbb{E}\left[ x_0 \right] \mathbb{E}\left[ x_0 \right]^\top \right) \left( A^k \right)^\top + \sum_{i=0}^k A^i B R_w B^\top \left( A^\top \right)^i$$
As you've correctly identified,
$$k \to \infty \implies A^{k} \left( \mathbb{E}\left[ x_0 x_0^\top \right] - \mathbb{E}\left[ x_0 \right] \mathbb{E}\left[ x_0 \right]^\top \right) \left( A^k \right)^\top \to 0$$
For the remaining terms, we need some functional analysis facts.
- $\mathbb{R}^{m \times m}$ with the operator norm is a Banach space.
- The set of PSD matrices $\mathbb{S}^m_+ \subseteq \mathbb{R}^{m \times m}$ is also a Banach space.
- In a Banach space $X$, every absolutely convergent series converges in $X$.
Therefore,
$$\sum_{i=0}^\infty \left \lVert A^i B R_w B^\top \left( A^\top \right)^i \right \rVert < \infty \implies \sum_{i=0}^{\infty} A^i B R_w B^\top \left( A^\top \right)^i \text{ converges in $\mathbb{S}_+^m$}$$
and it only remains for us to show $\sum_{i=1}^\infty \left \lVert A^i B R_w B^\top \left( A^\top \right)^i \right \rVert < \infty$.
$$\sum_{i=0}^\infty \left \lVert A^i B R_w B^\top \left( A^\top \right)^i \right \rVert \leq \left \lVert B R_w B^\top \right \rVert \sum_{i=0}^\infty \lambda_{\max}\left( A \right)^{2i} < \infty$$