Briefly, it's "safe" to apply an injective (one-to-one) function whose domain is $\mathbb{R}$ to both sides of an equation. This condition is generally too strong, see below, but I think it properly addresses the original question.
In more detail, say your original equation has the form $A=B$, in which $A$ and $B$ are (for simplicity, real-valued) expressions in one or more variables. After transformation, your equation has the form $f(A)=f(B)$ for some function $f$. The question is: "When is the solution set of $f(A)=f(B)$ identical to that of $A=B$?" Injectivity, by definition, means that "$f(A)=f(B)$ implies $A=B$". (The converse implication is built into the definition of a function.)
In the original question, $f$ was "squaring", $f(A)=A^2$, which of course is not injective on $\mathbb{R}$. "Safe" choices of $f$ might represent adding an expression, multiplying by a non-zero expression, extracting an odd root, or exponentiation.
Once this idea makes sense, it's easy to extend. For example, if you know that $A$ and $B$ represent positive real numbers for all values of the variables, it's "safe" to take (real) logarithms or (non-negative) square roots of both sides. If you know $|A|$ and $|B|$ are no larger than $\pi/2$, it's "safe" to take the sine of both sides, etc., etc.
Two tangential remarks:
One reason increasing functions are useful is that they preserve inequalities: $A<B$ if and only if $f(A)<f(B)$. Similar statements for decreasing and non-decreasing functions are left as an exercise.
Over the complex numbers, exponentiation isn't injective, so log has branches, which leads to all manner of fun. My favorite example (found by a classmate many years ago) is this short "proof":
$$e^z = (e^{2\pi i})^{z/2\pi i} = 1^{z/2\pi i} = 1\quad\text{for all complex $z$}.$$
