I realized that $\vec{W}$, $\vec{Y}$ and $\vec{Z}$ depend on both $\vec{r}$ and $t$ independently, anot not through a composite $t_r$. So here is an enhanced version of the answer
Let us say $\mathbb{V}_s \subset \mathbb{R}^3$ is the smallest spherical volume such that:
$$
\forall t, \qquad \forall \vec{r}\notin \mathbb{V}_s: \qquad\qquad \rho\left(\vec{r},t\right) = \vec{J}\left(\vec{r},t\right) = 0
$$
$$
\forall t, \qquad \forall \vec{r}\in \partial\mathbb{V}_s: \qquad\qquad \rho\left(\vec{r},t\right) = \vec{J}\left(\vec{r},t\right) = 0
$$
Since $\mathbb{V}_s$ is a sphere, let us say that its center is at $\vec{r}_0$, and its diameter id $\mathcal{D}=2\mathcal{R}_s$.
Now given the definitions:
$$
\vec{r}_s \in \mathbb{V}_s,\qquad
\vec{R} = \vec{r} - \vec{r}_s,\qquad
R=\left|\vec{R}\right|, \qquad
\hat{R}=\frac {\vec{R}} {R}, \qquad
t_r = t - \frac {R} {c}
$$
The fields $\vec{E}$ and $\vec{B}$ are as follows:
$$
\vec{E}\left(\vec{r},t\right) =
\frac {1} {4 \pi \epsilon_0}
\iiint_{\mathbb{V}_s} {\left[
\frac {\rho (\vec{r}_s, t_r)} {R^2} \hat{R} +
\frac {1} {R c} \frac {\partial \rho (\vec{r}_s, t_r) } {\partial t} \hat{R} -
\frac {1} {R c^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right]}
\space dV\left(\vec{r}_s\right)
$$
$$
\vec{B}\left(\vec{r},t\right) =
\frac {\mu_0} {4 \pi}
\iiint_{\mathbb{V}_s} {\left[
\frac {\vec{J} (\vec{r}_s, t_r)} {R^2} \times \hat{R} +
\frac {1} {R c} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \hat{R} \right]}
\space dV\left(\vec{r}_s\right)
$$
The paper "The Relation Between Expressions for Time-Dependent Electromagnetic Fields Given by Jefimenko and by Panofsky and Phillips" by Kirk T. McDonald shows how the continuity equation $-\dot{\rho}=\nabla\cdot\vec{J}$ can be applied to the expression for $\vec{E}$ to transform it into:
$$
\scriptsize{
\vec{E}\left(\vec{r},t\right) =
\frac {1} {4 \pi \epsilon_0}
\iiint_{\mathbb{V}_s} {\left[
\frac {\rho (\vec{r}_s, t_r)} {R^2} \hat{R}
+ \frac
{ \left(\vec{J} (\vec{r}_s, t_r) \cdot \hat{R}\right)\hat{R} + \left(\vec{J} (\vec{r}_s, t_r) \times\hat{R}\right) \times \hat{R} }
{R^2 c}
+ \frac {1} {R c^2} \left( \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \hat{R} \right) \times \hat{R}
\right]}
\space dV\left(\vec{r}_s\right)
}
$$
Now given any $\vec{r}$, the maximum possible difference between the values of $\dfrac 1 R$ for any two points $\vec{r}_s \in \mathbb{V}_s$ is:
$$\frac{1}{R} - \frac{1}{R+\mathcal{D}} = \frac{\mathcal{D}}{R\left(R+\mathcal{D}\right)} = \frac{1}{R}\cfrac{ \color{red}{\frac{\mathcal{D}}{R}}}{1+\color{red}{\frac{\mathcal{D}}{R}}}$$
Similarly, the maximum possible angle between the values of $\hat{R}$ for any two points $\vec{r}_s \in \mathbb{V}_s$ is:
$$
2\;{\tan}^{-1}\left(\frac12 \color{red} {\frac{\mathcal{D}}{R}}\right)
$$
For values of $\vec{r}$ for which $R \gg \mathcal{D}$, we have $\dfrac {\mathcal{D}} R \approx 0$, and each of the two expressions above are $\approx 0$ as well. So $\dfrac 1 R$ and $\hat{R}$ become essentially independent of $\vec{r}_s$. We can, therefore, bring out all $\dfrac 1 R$ and $\hat{R}$ factors outside the integral sign.
So if we define the following:
$$
W\left(\vec{r},t\right) =
\iiint_{\mathbb{V}_s}
\rho \left( \vec{r}_s, t_r \right)
\space dV\left(\vec{r}_s\right)
$$
$$
\vec{Y}\left(\vec{r},t\right) =
\iiint_{\mathbb{V}_s}
\vec{J} \left( \vec{r}_s, t_r \right)
\space dV\left(\vec{r}_s\right)
$$
$$
\vec{Z}\left(\vec{r},t\right) =
\iiint_{\mathbb{V}_s}
\frac {\partial \vec{J} \left( \vec{r}_s, t_r \right) } {\partial t}
\space dV\left(\vec{r}_s\right)
= \frac {\partial \vec{Y}\left(\vec{r},t\right)} {\partial t}
$$
we can then say:
$$
\vec{E}\left(\vec{r},t\right) =
\frac {1} {4 \pi \epsilon_0}
\left[
\frac {W (\vec{r}, t)} {R^2} \hat{R}
+ \frac {\vec{Y} (\vec{r}, t) \cdot \hat{R}} {R^2 c} \hat{R}
+ \frac {\left(\vec{Y}(\vec{r}, t) \times \hat{R}\right) \times \hat{R}} {R^2 c}
+ \frac {\left(\vec{Z}(\vec{r}, t) \times \hat{R}\right) \times \hat{R}} {R c^2}
\right]
$$
$$
\vec{B}\left(\vec{r},t\right) =
\frac {\mu_0} {4 \pi}
\left[
\frac {\vec{Y} (\vec{r}, t) \times \hat{R}} {R^2}
+ \frac {\vec{Z}(\vec{r}, t) \times \hat{R}} {R c}
\right]
$$
Applying $\color{blue}{\left(\vec{a} \times \hat{u}\right) \times \hat{u} \equiv \left(\vec{a}\cdot\hat{u}\right)\hat{u} - \vec{a}}$, we get:
$$
\vec{E}\left(\vec{r},t\right) =
\frac {1} {4 \pi \epsilon_0} \left[
\left(
\frac {W (\vec{r}, t)} {R^2}
+ 2\frac
{\vec{Y} (\vec{r}, t) \cdot \hat{R}}
{R^2 c}
+ \frac
{\vec{Z}\left( \vec{r}, t \right)\cdot\hat{R}}
{R c^2}
\right) \hat{R}
- \frac 1 {Rc} \left(
\frac {\vec{Y} (\vec{r}, t) } {R}
+ \frac {\vec{Z}(\vec{r}, t)} {c}
\right)
\right]
$$
$$
\vec{B}\left(\vec{r},t\right) =
\frac {\mu_0} {4 \pi} \left[
\frac 1 R
{\left(
\frac {\vec{Y} (\vec{r}, t)} {R} +
\frac {\vec{Z}(\vec{r}, t)} {c} \right)}
\times \hat{R}
\right]
$$
Thus, if we define:
$$
\mathcal{U}\left(\vec{r},t\right) =
\color{blue}{
\vec{W} \left(\vec{r},t\right)
+ \frac {\vec{Y} \left(\vec{r},t\right) \cdot \hat{R}} {c}
}
= \iiint_{\mathbb{V}_s} \left[
\rho \left( \vec{r}_s, t_r \right)
+ \frac { \vec{J} \left( \vec{r}_s, t_r \right) \cdot \hat{R} } {c}
\right] \space dV\left(\vec{r}_s\right)
$$
$$
\vec{\mathcal{X}} \left(\vec{r},t\right) =
\color{blue}{
\frac {\vec{Y} \left(\vec{r},t\right)} {R}
+ \frac {\vec{Z} \left(\vec{r},t\right)} {c}
}
= \iiint_{\mathbb{V}_s} \left[
\frac { \vec{J} \left( \vec{r}_s, t_r \right) } {R}
+ \frac 1 c \frac {\partial \vec{J} \left( \vec{r}_s, t_r \right) } {\partial t}
\right] \space dV\left(\vec{r}_s\right)
$$
... we get:
$$
\vec{E}\left(\vec{r},t\right) =
\frac {1} {4 \pi \epsilon_0}
\left[
\frac {\mathcal{U}\left(\vec{r},t\right)} {R^2} \hat{R}
+ \frac {\left( \vec{\mathcal{X}}\left(\vec{r},t\right) \times \hat{R}\right) \times \hat{R}} {Rc}
\right]
$$
$$
\vec{B}\left(\vec{r},t\right) =
\frac {\mu_0} {4 \pi}
\left[
\frac {\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R}} {R}
\right]
$$
Finally we can compute the Poynting vector $\vec{S}\left(\vec{r},t\right) = \dfrac {\vec{E}\left(\vec{r},t\right) \times \vec{B}\left(\vec{r},t\right)} {\mu_0}$ giving us:
$$
\vec{S}\left(\vec{r},t\right) =
\mathcal{U} \left(\vec{r},t\right) \frac {\hat{R} \times \left( \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right) } {16 \pi^2 \epsilon_0 R^3}
+ \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 R^2 c} \hat{R}
$$
The first term is clearly perpendicular to $\hat{R}$, while the second term is along $\hat{R}$. If we define a sphere $\mathbb{V}$ with its center at $\vec{r}_0$ (the center of the source volume $\mathbb{V}_s$) such that point $\vec{r}$ is on its surface, then the unit vector $\hat{R}$ will actually be the unit normal (to the surface $\partial\mathbb{V}$, pointing outward) at $\vec{r}$.
The outward power flow through $\vec{r}$ is given by:
$$
\vec{S}\left(\vec{r},t\right) \cdot \hat{R} = \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 R^2 c}
$$
We can see that $\vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R}$ is perpendicular to $\hat{R}$, and is therefore tangential to the surface $\partial\mathbb{V}$ everywhere. By the Hairy Ball Theorem, it must be zero for at least one $\vec{r}$. The outward power flowing through $\partial\mathbb{V}$ must therefore be zero at at least one point (and not one axis, or two points, as was suspected in the question).
The total power passing the surface $\partial\mathbb{V}$ will be:
$$
P\left(R,t\right) = \oint_{\partial\mathbb{V}}
\vec{S}\left(\vec{r},t\right) \cdot \hat{R}
\;ds\left(\vec{r}\right)
= \oint_{\partial\mathbb{V}}
\frac
{\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2}
{16 \pi^2 \epsilon_0 R^2 c}
\;ds\left(\vec{r}\right)
$$
If we now express the vectors in spherical coordinates $\langle 0\le \mathrm{r} \lt \infty, \;0 \le \theta \le 2\pi,\; 0 \le \phi \le \pi\rangle$ with the origin at $\vec{r}_0$, this surface integral can be parameterized to become:
$$ \small{
P\left(R,t\right)
= \int_0^{2\pi} \int_0^{\pi} \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 R^2 c} R^2 \sin \phi \;d\phi\;d\theta
= \int_0^{2\pi} \int_0^{\pi} \frac {\left| \vec{\mathcal{X}} \left(\vec{r},t\right) \times \hat{R} \right|^2} {16 \pi^2 \epsilon_0 c}\sin \phi \;d\phi\;d\theta
} $$
... which expands to:
$$\small{ \frac 1 {16 \pi^2 \epsilon_0 c} \int_0^{2\pi} \int_0^{\pi} \left[ \frac {\left| \vec{Y} \left(\vec{r},t\right) \times \hat{R} \right|^2} {R^2}
+ 2\frac {\left( \vec{Y} \left(\vec{r},t\right) \times \hat{R} \right)\cdot\left( \vec{Z} \left(\vec{r},t\right) \times \hat{R} \right)} {R c}
+ \frac {\left| \vec{Z} \left(\vec{r},t\right) \times \hat{R} \right|^2} {c^2}
\right]\;\sin \phi \;d\phi\;d\theta }$$
As $R \to\infty$, we can see that the first two terms vanish, and the term $\mathcal{P}_{\infty}$ from the question turns out to be:
$$ \small{
\mathcal{P}_{\infty}\left(t\right)
= \lim_{R\to\infty} P\left(R,t+ \frac R c\right)
= \frac 1 {16 \pi^2 \epsilon_0 c^3} \int_0^{2\pi} \int_0^{\pi} \left| \color{red}{ \vec{\mathcal{Q}}\left(\hat{R}, t\right) } \times \hat{R} \right|^2\;\sin \phi \;d\phi\;d\theta }
$$
... where:
$$
\vec{\mathcal{Q}}\left(\hat{R}, t\right)
= \lim_{R\to\infty}\vec{Z} \left(\vec{r},t+\frac R c\right)
= \iiint_{\mathbb{V}_s} \frac {\partial \vec{J} \left( \vec{r}_s, t+\frac {\left(\vec{r}_s-\vec{r}_0\right)\cdot\hat{R}} c\right) } {\partial t} \space dV\left(\vec{r}_s\right)
$$
$\vec{\mathcal{Q}}$ depends on the direction $\hat{R}$, but is completely independent of the distance $R$.
