In order to get a bit more insight into this it helps to consider the two extreme end cases. Namely, when $Q=0$ or when $R=0$. When doing this it also helps to know how your algebraic Riccati equation for $P_\infty$ can be derived and what that matrix actually represents. For this I will consider the following more general model
\begin{align}
x_{k} &= F_k\,x_{k-1} + B_k\,u_k + w_k, \quad w_k \sim \mathcal{N}(0,Q_k) \tag{1} \\
z_k &= H_k\,x_k + v_k, \quad v_k \sim \mathcal{N}(0,R_k) \tag{2}
\end{align}
where $u_k$ is deterministic and known, and $v_k$ and $w_k$ are uncorrelated. The Kalman filter equations for this model are given by
\begin{align}
\hat{x}_{k|k-1} &= F_k\,\hat{x}_{k-1|k-1} + B_k\,u_k, \tag{3} \\
P_{k|k-1} &= F_k\,P_{k-1|k-1}\,F_k^\top + Q_k, \tag{4} \\
K_k &= P_{k|k-1}\,H_k^\top (H_k\,P_{k|k-1}\,H_k^\top + R_k)^{-1}, \tag{5} \\
\hat{x}_{k|k} &= \hat{x}_{k|k-1} + K_k\,(z_k - H_k\,\hat{x}_{k|k-1}), \tag{6} \\
P_{k|k} &= (I - K_k\,H_k) P_{k|k-1}, \tag{7}
\end{align}
with $\hat{x}_{k|j}$ the expected value of $x_{k}$ given the measurements up to and including $z_j$ and $P_{k|j}$ the covariance of $\hat{x}_{k|j}$, i.e. $\mathbb{E}[(x_{k}-\hat{x}_{k|j})(x_{k}-\hat{x}_{k|j})^\top]$. Here is can be noted that $P_{k|k-1}$ means something else than $P_{k|k}$ and in most cases these are not equal to each other, also not in the limit case as $k\to\infty$.
In order to obtain a discrete time algebraic Riccati equation in the covariance, one must substitute $(4)$ and $(7)$ into each other. For example incrementing $k$ by one in $(4)$ and substituting in $(7)$ using $(5)$, which yields
$$
P_{k+1|k} = F_{k+1}\,P_{k|k-1}\,F_{k+1}^\top - F_{k+1}\,P_{k|k-1}\,H_k^\top (H_k\,P_{k|k-1}\,H_k^\top + R_k)^{-1}\,H_k\,P_{k|k-1}\,F_{k+1}^\top + Q_{k+1}. \tag{8}
$$
This simplifies to your discrete time algebraic Riccati equation when using $P_{k+1|k} = P_{k|k-1} = P_\infty$, $F_k=H_k=I$, $R_k=R$ and $Q_k=Q$. If instead one wants to have the limit of $k\to\infty$ of $P_{k|k}$ (assuming that $F_k$, $H_k$, $Q_k$ and $R_k$ are constant), one could formulate another algebraic Riccati equation, or just apply $(7)$ once to the solution obtained from solving $(8)$. In order to distinguish between the limit of $P_{k|k-1}$ and $P_{k|k}$ I will denote $k\to\infty$ of $P_{k|k-1}$ as $P_\infty$ and $k\to\infty$ of $P_{k|k}$ as $P_\infty^*$.
Since $F_k$, $H_k$, $Q_k$ and $R_k$ are assumed constant in order to solve for $P_\infty$ and $P_\infty^*$, I will just drop the $k$ subscript, which simplifies $(8)$ to the following algebraic Riccati equation in $P_\infty$
$$
P_\infty = F\,P_\infty\,F^\top - F\,P_\infty\,H^\top (H\,P_\infty\,H^\top + R)^{-1}\,H\,P_\infty\,F^\top + Q. \tag{9}
$$
For the limit case of $Q=0$ it can be shown using $(9)$ that $P_\infty=0$ is a solution. Similarly using $(7)$ it also follows that $P_\infty^*=0$. It can be noted that $Q=0$ means that the dynamics from $(1)$ is actually deterministic and only the measurements $z_k$ are stochastic. This can also be seen as a least squares problem and the more and more noisy data (the measurements $z_k$) are used to fit the error gets smaller and smaller until it is zero, also see recursive least squares. Using this solution to obtain the Kalman gain from $(5)$ gives that this gain is also zero. Note that this is only the gain as $k\to\infty$, denoted with $K_\infty$. However, using $(3)-(7)$ starting with a non-zero $P_{k|k-1}$ will initially yield non-zero $K_k$, but the gain tends to zero over time.
In short it can be summarized that for the case of $Q=0$ the Kalman filter is more and more relying on the state predictions due to the deterministic dynamics from $(1)$ and in the end not using the measurements at all, i.e. $K_\infty=0$.
For the limit case of $R=0$ together with $H=I$ it can be shown using $(9)$ that $P_\infty=Q$. Similarly using $(7)$ it follows that $P_\infty^*=0$. I will start with the result of $P_\infty^*=0$, which can be understood using that the measurement $z_k$ measures the complete state $x_k$ without noise. So the best estimate of the state given such measurement is just the measurement itself. Why it should hold that $P_\infty^*=0$ can be seen from that $P_\infty^*$ is the limit of $P_{k|k}$, the covariance of $\hat{x}_{k|k}$, with $\hat{x}_{k|k} = z_k = x_k$. From $P_\infty^*=0$ the result for $P_\infty$ can be derived using $(4)$. Namely, $P_\infty$ is the limit of $P_{k|k-1}$, so the covariance of the state estimate predicted one time step into the future. Using this solution to obtain the Kalman gain from $(5)$ gives that $K_\infty = I$, which combined with $H=I$ simplifies $(6)$ to the expected result of $\hat{x}_{k|k} = z_k$.
In short it can be summarized that for the case of $R=0$, together with $H=I$, the Kalman filter is fully relying on the noise free measurements of the full state.
When $Q\neq0$ and $R\neq0$ then each time step the Kalman filter will not only really on only prediction steps or only measurements, but a weighted middle road between the two. It can be noted asking the meaning of $P_\infty=R$ is not a very meaningful question, because the left and right hand side represent the covariances of different variables with different units. For example one could always transform that state using $q_k = T\,x_k$ and obtain an equivalent state space model.
However, some insights might be gained when using your model with $H = I$ and $F=I$, and setting $P_\infty = \alpha\,R$, with $\alpha$ some positive scalar. Using this and solving $(9)$ for $Q$, $(5)$ for $K$ and $(7)$ for $P_\infty^*$ yields
$$
Q = \frac{\alpha^2}{1+\alpha} R, \\
K = \frac{\alpha}{1+\alpha} I, \\
P_\infty^* = \frac{\alpha}{1+\alpha} R.
$$
When setting $\alpha=0$ obtains the solution of when $Q=0$, so the Kalman filter is relying more and more on the prediction step. When $\alpha\to\infty$ one gets $Q=\alpha\,R$, $K=I$ and $P_\infty^*=R$. For which it can be noted that this is approaching the solution for $R=0$, but with all covariance matrices scaled by $\alpha$, so the Kalman filter is relying more and more only on the measurements, since the dynamics is subjected to way too much noise, making the predictions step unreliable.
It is worth noting that for all $\alpha$ it holds that $P_\infty \geq Q$ and $P_\infty^* \leq R$, but $P_\infty$ can be large or smaller than $R$ and $P_\infty^*$ can be large or smaller than $Q$. The constraint on $P_\infty$ can be understood by looking at $(4)$, since $P_{k-1|k-1}\geq0$ so $P_{k|k-1} \geq Q$, or in other words the covariance of the state estimate after one prediction step always is affected by the uncertainty of the prediction step itself. The constraint on $P_\infty^*$ can intuitively be understood by the fact that if $P_\infty^* > R$ then one would obtained a better state estimate by just taking the measurement as new state estimate, which would have covariance of $R$.
As far as I know there is no meaningful intuition for your case with $\alpha=1$. One thing that might be relevant is instead of using the Riccati equation from $(9)$ start with the Kalman equation starting at $(5)$ using $P_{k|k-1} = R$. This gives $K_k = I/2$, so in $(6)$ the corrected state estimate is the average of the prediction step and the measurement. Both the prediction step and the measurement have independent distributions but with the same variance, which could then also intuitively explain why $(7)$ gives $P_{k|k}=P_{k|k-1}/2=R/2$. Now the only remaining step is using $(4)$ and in order to obtain $P_{k|k-1}=R$ requires $Q=R/2$.
This answer might not have given a very clear explanation of the relation between $R$ and $Q$ when $P_\infty=R$. But I hope that the explanations for the cases of $Q=0$ and $R=0$ did give you more insights into the stationary solutions (the solution obtained by solving the algebraic Riccati equation) of the Kalman filter.
