According to my book: $$\ln(\lvert y \rvert)=\ln(\lvert 1+x \rvert)+ \ln(\lvert c \rvert)$$ $$\Rightarrow \ln(\lvert y \rvert)=\ln(\lvert c(1+x) \rvert)$$ $$\Rightarrow y=c(1+x)$$
I know that this is correct. However, I want to be able to understand the second step clearly. I want to make sure if this is why : $$\ln(\lvert y \rvert)=\ln(\lvert c(1+x) \rvert) \Rightarrow \lvert y \rvert=\lvert c(1+x) \rvert$$ $$\Rightarrow y=± c(1+x) \Rightarrow y=C(1+x)$$ where $C=±c$ is just a substitution for the constant. Am I correct? Is this an often-used method that is considered rigorous even if we do not rename the constant?
Edit: $c$ is not a fixed constant, but an arbitrary one. To give some context, the equation to be solved is obtained from integration.
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.)