$$u_t+u\:u_x=-cu\tag 1$$
$$ds=\frac{dt}{1}=\frac{dx}{u}=\frac{du}{-c}$$
A first characteristic equation comes from solving $\frac{dt}{1}=\frac{du}{-c}$ :
$$u\,e^{ct}=c_1$$
A second characteristic equation comes from $\frac{dx}{u}=\frac{du}{-c}$:
$$u+c\,x=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ with arbitrary function $F$ is :
$$\boxed{u+c\,x=F\left(u\,e^{ct}\right)}\tag 2$$
CONDITION :
$u(x,0)=u_0(x)$ with known (given) function $u_0(x)$ .
$$u_0(x)+c\,x=F\left(u_0(x)\right)$$
Consider the inverse function of $X=u_0(x)$ say $x(X)=u_0^{(-1)}(X)$
The superscript $(-1)$ denotes the inverse function, not a power $-1$. Take care to not confuse.
$$X+c\,u_0^{(-1)}(X)=F\left(X\right)$$
Since $u_0(x)$ is a known function we can consider that the inverse function $u_0^{(-1)}(X)$ is known in general. Thus the function $F(X)$ is known now :
$$F\left(X\right)=X+c\,u_0^{(-1)}(X)$$
We put it into the above general solution $(2)$ where $X=u\,e^{ct}$ :
$$\boxed{u+cx=u\,e^{ct}+c\,u_0^{(-1)}\left(u\,e^{ct}\right)}\tag 3$$
In this implicit equation, $u(x,t)$ is the particular solution of the PDE which agrees with the specified condition.
When it is possible, solving the implicit equation for $u$ would give the solution on explicit form. Of course this is not possible if $u_0(x)$ is not given explicitly. But even if $u_0(x)$ is explicitly given it is not always possible with standard functions. In those cases one consider the implicit solution $(3)$ as the final solution of the problem.
