Yes, you are correct. Though I would not write $C=\pm c$. Note that $c$ is a fixed (nonzero) constant. You cannot define $C$ with two values.
In general, what you have is that
$$
|y|=|kx|
$$
if and only if
$$
y=kx\quad \textrm{or}\quad y=-kx\tag{1}
$$
(1) is sometimes written as $y=\pm kx$.
Notes 1.
If you get the identity when you solve an ODE and get two (sets of) solutions: $y=cx$ and $y=-cx$, you need to be careful if you want to combine them. In your case, the "arbitrary" constant $c$ is not zero. So you cannot conclude that $y=Cx$ where $C$ is "arbitrary". If you find further that $y=0$ is also a solution, then you can indeed claim that $y=Cx$, combining all the cases.
Notes 2.
Consider for instance the ODE you mentioned in the comment:
$$
\frac{dy}{dx}=\frac{y}{1+x}\tag{0}
$$
To solve it,
\begin{align}
\frac{dy}{y}&=\frac{dx}{1+x}\\
\ln|y|&=\ln|1+x|+\ln|c|&(c\ne 0)
\end{align}
(Remark. The reason why we can write $\ln|c|$ here is that the range of the function $x\mapsto \ln x$ ($x>0$) is the whole real line. So adding an "arbitrary" constant $c$ is the same as adding $\ln|c|$ for some arbitrary nonzero constant $c$.)
To go on,
\begin{align}
\ln|y|&=\ln|1+x|+\ln|c|&(c\ne 0)\\
\ln|y|&=\ln|c(1+x)|&(c\ne 0)\\
|y|&=|c(1+x)|&\textrm{(because $\ln x$ is injective)}
\end{align}
Now since $c$ is an arbitrary nonzero constant, you have following set of solutions,
$$
y=c(1+x),\quad c\ne 0.\tag{1}
$$
Since $c$ is "arbitrary", you don't need to add $\pm$ in front of it.
Observe that $y\equiv 0$ is also a solution to (0), and it is not of the form in (1).
So combining (1) and the zero solutions together, you get that the general solution for (0) is
$$
y=C(1+x)
$$
where $C$ is an arbitrary constant. (Note that $C$ is not required to be nonzero anymore.)
