You are very close, but slightly off, and I think the confusion comes in this variable of integration $r$.
I'll present the exposition that has always made most sense to me. Characteristics $(x(\tau),t(\tau))$ originating from a point $(x(0),t(0)) = (\xi,0)$ are given by \begin{align*}
\dot x &= b, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x(0) = \xi, \\
\dot t &= 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, t(0) = 0, \\
\dot z &= s(x,t), \,\,\, z(0) = u_0(\xi),
\end{align*} where $z(\tau;\xi) = u(x(\tau;\xi),t(\tau;\xi))$. Solving these gives \begin{align*}
x(\tau;\xi) &= b\tau+\xi, \\
t(\tau;\xi) &= \tau, \\
z(\tau;\xi) &= u_0(\xi) + \int^\tau_{0} s(x(r;\xi),t(r;\xi))dr = u_0(\xi) + \int^\tau_{0} s(br+\xi,r)dr.
\end{align*} Now if we can invert the relationships $(x(\tau;\xi),t(\tau;\xi))$ to arrive at $(\tau(x,t),\xi(x,t))$, then we have the solution $u(x,t) = z(\tau(x,t);\xi(x,t))$. Note that the integration variable $r$ will stay! Only $\tau$ and $\xi$ will be replaced. As you've noted $\xi = x-bt$, and we also have $\tau = t$. Thus \begin{align*} u(x,t) &= u_0(x-bt)+\int^t_0 s(br+x-bt,r)dr \\
&= u_0(x-bt)+\int^t_0 s(x-b(t-r),r)dr.
\end{align*} We can check that this is indeed a solution. Note that when you differentiate the integral term with respect to $t$, you need to use the Leibniz rule. We see: \begin{align*} u_t(x,t) &= -bu'_0(x-bt) + s(x,t) -b \int^t_0 s_x(x-b(t-r),r)dr,\\
u_x(x,t) &= u'_0(x-bt) + \int^t_0 s_x(x-b(t-r),r)dr.
\end{align*} Thus we do indeed have $$u_t(x,t)+bu_x(x,t) = s(x,t),$$ and clearly $u(x,0) = u_0(x)$.
You are close. Parameterize the line $t=0$ by the initial position parameter $x_0$. Then the characteristic equation for $x$ reads
$$
\frac{dx}{dt} =1, \quad x(0) = x_0 \implies x(t) = t + x_0.
$$
Now the characteristic equation for $u$ reads
$$
\begin{aligned}
\frac{du(x(t),t)}{dt} &= \sin(x(t) - t) = \sin(x_0) \\
u(x(0,0)) &= \sin(x_0).
\end{aligned}
$$
Notice that everything involving $x_0$ here is independent of $t$; $x_0$ is just an independent parameter that describes where a single characteristic trajectory starts. Solving this ODE yields $u(x(t),t) = (t+1)\sin(x_0)$. To get things in terms of just $x$ and $t$, we solve for $x_0$ and substitute. Indeed, $x_0 = x-t$, so we have $u(x,t) = (1+t)\sin(x-t)$, as desired.
Best Answer
Your constant of integration is wrong. The characteristic going through $(x,t)$ hits the axis $t=0$ at $x_0=x+\cos t-1$ and your solution is $u(x,t)=(x+\cos t-1)^2$