The radial distribution has a different form due to integration over the angles:
If we take the absolute square of the wave function $\Psi$ and integrate over the whole volume, we get the Norm of the wave function
\begin{equation}
N = \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ \int \limits _{r=0}^{\infty } |\Psi(r,\theta ,\varphi )|^2r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi
\end{equation}
where $r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi$ comes from transforming the infinitesimal volume element from cartesian coordinates ($\mathrm{d} V = \mathrm {d} x\,\mathrm {d} y\,\mathrm {d} z$) to spherical coordinates.
For the radial probability distribution $P(r)$ we do the same, but omit the integral over $r$
\begin{equation}
P(r) = \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ |\Psi(r,\theta ,\varphi )|^2r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi
\end{equation}
which we can rearrange to
\begin{equation}
P(r) = |R(r)|^2r^{2}\int \limits _{\theta =0}^{\pi }\int \limits _{\varphi =0}^{2\pi }|Y(\theta ,\varphi)|^2\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi
\end{equation}
Where $R(r)$ is the radial part of the wave function $\Psi(r,\theta ,\varphi)=R(r)Y(\theta ,\varphi)$.
For the $1s$ orbital we $Y(\theta ,\varphi)=1$ and straight forward integration over $\theta$ and $\varphi$ yields
\begin{equation}
P(r) = |R(r)|^2r^{2} 4\pi
\end{equation}
So no matter what orbital $\Psi$ represents, $P(r)$ always has a factor $r^2$, which means we always have $P(r=0)=0$.
Your given example is the $1s$ orbital $\Psi_{1s}\propto \exp(-r)$. In contrast to all other orbitals, it does not have a polynomial in $r$. Therefore it has no node at $r=0$.
