Let $X$ be a set and $\sim$ be an equivalence relation on $X$. Then the set $$X/\sim{}:=\{[x] \ | \ x \in X\}$$ is the set of all equivalence classes with $[x]:= \{y \in X \ | \ y \sim x\}$. It is now possible to define sets such as $Y \subseteq X/\sim{}$ via some property $P$, for example $$Y:=\{[y] \ | \ P(y), y \in X\}.$$ If this property $P$ is not independent of the choice of representation, this would lead to contradictions regarding set membership and is a well defined set. If, however, $P$ is a property that is independent, why does this not lead to any contradictions at all? What I mean is, what guarantees, that there can't be any contradictions whenever $P$ is an independent property.
An easy example might be $Y \subseteq \mathbf{Z}/2\mathbf{Z}$ with $Y:=\{[z] \ | \ z \ \mathrm{even} \ \}.$
The purpose isn't to avoid a contradiction, which we presume not to be an issue since we presume our set theory to be consistent. It's just to avoid behaviors we don't want. Notice that the set you call $Y$ exists by the ZF axioms no matter what $X,\sim,P$ are, so ZF (and much weaker theories) would be inconsistent if we could get a contradiction from choosing any combination of these.
To see why there's no inconsistency if the predicate does not respect the equivalence relation, define the equivalence relation $m\sim n$ on $\mathbb{N}$ to be $m=m$; that is, it's the total equivalence relation. Now consider $Y=\{[n]\;|\;n\in\mathbb{N}, n\textrm{ is even}\}$. This is a perfectly good set that poses no logical problems. Why?
Note that in the definition of $Y$, $n$ is actually bound. So what is it bound by? The answer is there's a suppressed existential quantifier when you start unpacking the set builder notation, because what it means for $Y$ to have this definition is that $$x\in Y\Leftrightarrow\exists n(x=\{z\in\mathbb{N}\;|\;n\sim z\}\wedge n\in\mathbb{N}\wedge n\textrm{ is even}).$$ So $Y$ will contain our single equivalence class, even though our condition "$n\textrm{ is even}$" certainly doesn't respect the equivalence relation, because "$n\textrm{ is even}$" is true of some $n$ in the equivalence class.
Edit:
To elaborate a little more, your worry only holds if it's true (as it appears you're tacitly supposing in your comments) that $$P(x)\Leftrightarrow [x]\in Y,$$ but this is actually equivalent to the supposition that $P$ respects the equivalence relation. For suppose that this biconditional is true, that $x\sim y$, and $[x]\in Y$; then $[y]\in Y$, so our biconditional tells us that $P(y)$. Hence under this assumption, $$P(x)\wedge x\sim y\Rightarrow P(y)$$ holds. That invariance of $P$ under $\sim$ implies $P(x)\Leftrightarrow [x]\in Y$ is trivial.
So this doesn't directly answer your question (which becomes, ultimately, whether our set theory is consistent), but I hope addressing the misconception in the question makes that seem less pressing.