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!
I found the answer to the question: "How do Rankine-Hugoniot conditions look in the case of balance laws?"
It is in the book: Rational extended thermodynamics - Muller, Ruggeri, 1998
In Chapter 8 - Subsection 5.1 (Weak solutions) and Subsection 5.2 (Rankine-Hugoniot Equations). It is given in general terms but it is the proof I was looking for.
So the Rankine-Hugoniot condition stays the same in the case (2),(3) as in the case (1),(3).
But what happens to the speed?
In the conservation law case (1),(3) shock speed is constant. For the balance law case (2),(3), I think that we calculate speed in the same way but it isn't constant anymore. I came to this conclusion by solving Burgers equation with and without source with some concrete functions $f,g$ and concrete constants $u_l,u_r$. But I am not sure that could be generalized. If I am wrong about this, please correct me.
If I get any new info I'll post it here.
Best Answer
If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:
The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e. $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x \leq t \\ &\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\ &0 & &\text{if}\quad x\geq 1 \end{aligned}\right. $$ At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$. Therefore, after the breaking time $t\geq 1$, $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\ &0 & &\text{if}\quad x > 1 +\tfrac12 (t-1) \end{aligned}\right. $$
To your questions: