Here I would like to expand on some of the arguments given in Ron Maimon's nice answer.
i) Setting. Let us divide the 1D $x$-axis into three regions $I$, $II$, and $III$, with a localized potential $V(x)$ in the middle region $II$ having a compact support. (Clearly, there are physically relevant potentials that haven't compact support, e.g. the Coulomb potential, but this assumption simplifies the following discussion concerning the notion of asymptotic states.)
ii) Time-independent and monochromatic. The particle is free in the regions $I$ and $III$, so we can solve the time-independent Schrödinger equation
$$\begin{align}\hat{H}\psi(x) ~=~&E \psi(x), \cr
\hat{H}~=~& \frac{\hat{p}^2}{2m}+V(x),\qquad E> 0,\end{align} \tag{1}$$
exactly there. We know that the 2nd order linear ODE has two linearly independent solutions, which in the free regions $I$ and $III$ are plane waves
$$ \begin{align} \psi_{I}(x) ~=~& \underbrace{a^{+}_{I}(k)e^{ikx}}_{\text{incoming right-mover}} + \underbrace{a^{-}_{I}(k)e^{-ikx}}_{\text{outgoing left-mover}},
\qquad k> 0, \tag{2} \cr
\psi_{III}(x) ~=~& \underbrace{a^{+}_{III}(k)e^{ikx}}_{\text{outgoing right-mover}} + \underbrace{a^{-}_{III}(k)e^{-ikx}}_{\text{incoming left-mover}}. \tag{3} \end{align} $$
Just from linearity of the Schrödinger equation, even without solving the middle region $II$, we know that the four coefficients $a^{\pm}_{I/III}(k)$ are constrained by two linear conditions. This observation leads, by the way, to the time-independent notion of the scattering $S$-matrix and the transfer $M$-matrix
$$ \begin{pmatrix} a^{-}_{I}(k) \\ a^{+}_{III}(k) \end{pmatrix}~=~ S(k) \begin{pmatrix} a^{+}_{I}(k) \\ a^{-}_{III}(k) \end{pmatrix},
\tag{4} $$
$$ \begin{pmatrix} a^{+}_{III}(k) \\ a^{-}_{III}(k) \end{pmatrix}~=~ M(k) \begin{pmatrix} a^{+}_{I}(k) \\ a^{-}_{I}(k) \end{pmatrix},
\tag{5} $$
see, e.g. my Phys.SE answer here.
iii) Time-dependence of monochromatic wave. The dispersion relation reads
$$ \frac{E(k)}{\hbar} ~\equiv~\omega(k)~=~\frac{\hbar k^2}{2m}.
\tag{6}\label{eq:6} $$
The specific form on the right-hand side of the dispersion relation $\eqref{eq:6}$ will not matter in what follows (although we will assume for simplicity that it is the same for right- and left-movers). The full time-dependent monochromatic solution in the free regions $I$ and $III$ becomes
$$\begin{align} \Psi_r(x,t)
~=~& \sum_{\sigma=\pm}a^{\sigma}_r(k)e^{\sigma ikx-i\omega(k)t}\cr
~=~&\underbrace{e^{-i\omega(k)t}}_{\text{phase factor}}
\Psi_r(x,0), \qquad
r ~\in~ \{I, III\}.\end{align} \tag{7}\label{eq:7} $$
The solution $\eqref{eq:7}$ is a sum of a right-mover ($\sigma=+$) and a left-mover ($\sigma=-$). For now, the words right- and left-mover may be taken as semantic names without physical content. The solution $\eqref{eq:7}$ is fully delocalized in the free regions $I$ and $III$ with the probability density $|\Psi_r(x,t)|^2$ independent of time $t$, so naively, it does not make sense to say that the waves are right or left moving, or even scatter! However, it turns out that we may view the monochromatic wave $\eqref{eq:7}$ as a limit of a wave packet and obtain a physical interpretation in that way; see next section.
iv) Wave packet. We now take a wave packet
$$\begin{align} A^{\sigma}_r(k)~=~&0 \qquad \text{for} \qquad
|k-k_0| ~\geq~ \frac{1}{L}, \cr
\sigma~\in~&\{\pm\}, \qquad r ~\in~ \{I, III\},\end{align}\tag{8} $$
narrowly peaked around some particular value $k_0$ in $k$-space,
$$|k-k_0| ~\leq~ K, \tag{9}$$
where $K$ is some wave number scale so that we may Taylor expand the dispersion relation
$$\omega(k)~=~ \omega(k_0) + v_g(k_0)(k-k_0) + {\cal O}\left((k-k_0)^2\right), \tag{10}\label{eq:10} $$
and drop higher-order terms ${\cal O}\left((k-k_0)^2\right)$. Here
$$v_g(k)~:=~\frac{\mathrm{d}\omega(k)}{\mathrm{d}k}\tag{11}$$
is the group velocity. The wave packet (in the free regions $I$ and $III$) is a sum of a right- and a left-mover,
$$ \Psi_r(x,t)~=~ \Psi^{+}_r(x,t)+\Psi^{-}_r(x,t),
\qquad r ~\in~ \{I, III\},\tag{12} $$
where
$$\begin{align} \Psi^{\sigma}_r(x,t)~:=~& \int \mathrm{d}k~A^{\sigma}_r(k)e^{\sigma ikx-i\omega(k)t}\cr
~\approx~& e^{i(k_0 v_g(k_0)-\omega(k_0))t}
\int \mathrm{d}k~A^{\sigma}_r(k)e^{ ik(\sigma x- v_g(k_0)t)}\cr
~=~&\underbrace{e^{i(k_0 v_g(k_0)-\omega(k_0))t}}_{\text{phase factor}}
~\Psi^{\sigma}_r\left(x-\sigma v_g(k_0)t,0\right), \cr
\qquad\sigma~\in~&\{\pm\}, \qquad r ~\in~ \{I, III\}.\end{align} \tag{13}\label{eq:13}$$
The right- and left-movers $\Psi^{\sigma}$ will be very long spread-out wave trains of sizes $\geq \frac{1}{K}$ in $x$-space, but we are still able to identity via eq. $\eqref{eq:13}$ their time evolution as just
a collective motion with group velocity $\sigma v_g(k_0)$, and
an overall time-dependent phase factor of modulus $1$ (which is the same for the right- and the left-mover).
In the limit $K \to 0$, with $K >0$, the approximation $\eqref{eq:10}$ becomes better and better, and we recover the time-independent monochromatic wave,
$$ A^{\sigma}_r(k) ~\longrightarrow ~a^{\sigma}_r(k_0)~\delta(k-k_0)\qquad \text{for} \qquad K\to 0. \tag{14}$$
It thus makes sense to assign a group velocity to each of the $\pm$ parts of the monochromatic wave $\eqref{eq:7}$ because it can understood as an appropriate limit of the wave packet $\eqref{eq:13}$. The previous sentence is, in a nutshell, the answer to OP's title question (v3).
