It’s true! All the passages in the proof that you already know are reversible.
Edit:
Let $ [u] \stackrel{\text{df}}{=} {u_{+}}(p) - {u_{-}}(p) $ for all $ p \in \gamma_{0} = \{ (\xi(t),t) \mid t \in I \} $.
(R.H.S.): As $ \xi'(t) = \dfrac{[f(u)]}{[u]}(\xi(t),t) $ for all $ t \in I $, we have
\begin{align}
& [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) = 0 \\ \Longrightarrow \quad
& 0 = \int_{I}
\Big( [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) \Big) \cdot
\phi(\xi(t),t) ~
\mathrm{d}{t}.
\end{align}
Let $ v = (v_{1},v_{2}) = (1,- \xi'(t)) $. We can then write the previous formula as follows:
\begin{align}
0
& = \int_{I} ([f(u)] v_{1} + [u] v_{2}) \phi ~ \mathrm{d}{s} \\
& = \int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
\mathrm{d}{s} -
\int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~
\mathrm{d}{s}.
\end{align}
We have taken $ \phi \in {C_{c}^{1}}(\Omega) $, so $ \text{supp}(\phi) = \omega_{-} \cup (\gamma_{0} \cap \text{supp}(\phi)) \cup \omega_{+} $.
This is the crucial point: We are going to use the Divergence Theorem in the ‘non-standard’ way:
\begin{align}
\int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
\mathrm{d}{s}
& = \int_{\partial \omega_{+}}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~\mathrm{d}{s} \\
& = \iint_{\omega_{+}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t} +
\iint_{\omega_{+}}
\left(
\frac{\partial u}{\partial t} + \frac{\partial f(u)}{\partial x}
\right) \phi ~ \mathrm{d}{x} \mathrm{d}{t}.
\end{align}
Note that the last integral is $ 0 $ because our solution is classical outside the shock. We now have the following equations:
$$
\int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
\mathrm{d}{s}
= \iint_{\omega_{+}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t},
$$
and, similarly,
$$
- \int_{\gamma_{0} \cap \text{supp}(\phi)}
\Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~ \mathrm{d}{s}
= \iint_{\omega_{-}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t}.
$$
Summing up term by term:
\begin{align}
0
& = \int_{I} \Big( [f(u)] v_{1} + [u] v_{2} \Big) \phi ~ \mathrm{d}{s} \\
& = \iint_{\omega_{+} \cup \omega_{-}}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t} \\
& = \int_{\text{supp}(\phi)}
\left(
u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
\right) ~
\mathrm{d}{x} \mathrm{d}{t}.
\end{align}
This is exactly what we were supposed to prove.
Hope it helps!
Of course, one can go back to the general definition of weak solutions, as proposed in OP (see related post for complements). I will propose a solution which may be more in the spirit of the book (Example 10 p. 336). The derivatives of the discontinuous function $u(x,t) = H(x-ct)$ viewed as a distribution are
\begin{aligned}
u_x &= \delta(x-ct) , & u_t &= -c \delta(x-ct) ,\\
u_{xx} &= \delta'(x-ct) , & u_{tt} &= c^2 \delta'(x-ct) .
\end{aligned}
Hence, $u_{tt} = c^2 u_{xx}$ and $u$ is called a “weak” solution of the wave
equation.
Best Answer
Let us make the change of variable $(\xi, \tau) = (x-at,t)$, in other words $(x,t) = (\xi + a\tau,\tau)$, so that \begin{aligned} &\phi_t = \phi_\xi \xi_t + \phi_\tau \tau_t = \phi_\tau - a\phi_\xi \\ &\phi_x = \phi_\xi \xi_x + \phi_\tau \tau_x = \phi_\xi \, . \end{aligned} Thus, using Fubini's theorem and integration by parts, \begin{aligned} \iint_{\Bbb R\times\Bbb R_+} u\, (\phi_t + a \phi_x)\, \text d x\,\text d t &= \iint_{\Bbb R\times\Bbb R_+} u \phi_\tau\, \text d \xi\,\text d \tau \\ &= \iint_{\Bbb R_+\times\Bbb R} u \phi_\tau\, \text d \tau\,\text d \xi \\ &= \int_{\Bbb R}\left[u\phi\right]_{\tau\in\Bbb R_+}\text d\xi - \iint_{\Bbb R_+\times\Bbb R} \underbrace{u_\tau}_{u_t + au_x =0} \phi\, \text d \tau\,\text d \xi \\ &= -\int_{\Bbb R}\left.(u\phi)\right|_{t = 0}\text dx \, . \end{aligned} We have shown that the definition holds for all smooth $\phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there is a sign mistake in OP.
Note that $u$ does not need to be continuous to apply integration by parts as above (see Wikipedia article, §Extension to other cases, and Wikipedia article, §A concrete example).