Given the equation $$a^2=b^2,$$ solving for $a$ gives $$\sqrt {a^2} = \sqrt {b^2}\\ |a| = |b|\\ a = \pm |b|.$$ How to validly get rid of the absolute value sign on $b$ to get the correct answer $a = \pm b$?
Dividing $a = \pm |b|$ into cases
Case one: $a = |b|$
$b = a$ or $b = -a$
$b = \pm a$
Case two: $a = -|b|$
$b = (-a)$ or $b = -(-a)$
$b = \pm (-a)$
and making $a$ the subject (correct me if I'm wrong): $$a = \pm b \quad\text{or}\quad a = -(\pm b).$$ I'm aware that $\pm b \neq - (\pm b),$ so, how does the previous line imply that $a = \pm b$ ?
No, it is in fact true that $\pm b=- (\pm b),$ because $$y=\pm x$$ isn't a genuine equation but notational shorthand for $$y\in\{x,-x\}.$$
Even though my instinct is to continue from your first chunk of working by simply writing \begin{gather}a = &\pm|b|\\=&{\pm}(\pm b)\tag#\\=&{\pm}b,\end{gather} there is a caveat.
The intermediate expression $$\pm (\pm b)$$ is arguably ambiguous as to whether the top and bottom signs are meant to separately/independently correspond, i.e., is the expression meant to be understood as \begin{align}&{\pm}(\pm b)\\=&{+}(+b)\:\:\text{or}\:\:{-}({-}b)\\=&b,\end{align} or whether \begin{align}&{\pm} (\pm b)\\=&{+}(+b)\:\:\text{or}\:\:{+}({-}b)\:\:\text{or}\:\:{-}(+b)\:\:\text{or}\:\:{-}({-}b)\\=&{\pm}b.\end{align}
In $(\#)$ and the final line of this answer, I mean the second interpretation, however, typically, the expression $$(\pm \,a\mp b)$$ is interpreted the first way, as $$\pm(a-b).$$
Summarising: an expression containing multiple occurrences of $\pm$ and/or $\mp$ may be disambiguated from its context.
user1015917's answer is the cleanest, as it sidesteps all occurrences of
±.