What exactly is $\mathbb{Z_n}$?

To put it shortly: yes.

What isn't important is what $\mathbb{Z}_n$ "is" as a set. Rather, what's important is that all these different formulations of $\mathbb{Z}_n$ (equivalence classes of integers, select integers with a certain operation, a different set of integers with a certain operation) all give rise to an algebraic structure (groups, rings, fields, depending on what you're working with) which are all isomorphic.

Though, if you're saying that $\mathbb{Z}_n$ is equivalence classes, then the accepted notation would probably look more like $\mathbb{Z}_n = \{[n]_n, [n + 1]_n, \dots\}$, or perhaps even $\mathbb{Z}_n = \{n + n\mathbb{Z}, (n + 1) + n\mathbb{Z}, \dots\}$, depending on your background and your personal aesthetic concerns. $n$, after all, is an integer, not an equivalence class of integers.

As Qiaochu Yian says in his comment, the equivalence class definition is usually preferred, as this makes certain proofs much easier.

Below I explain the general idea, which works not only for rings but also for any algebraic structure definable by universal equational axioms ("identities") such as $\,x(y+z) = xy+xz\,$.

$a\equiv b\pmod{\! n}\iff n\mid a-b\,$ is a congruence relation (on a ring) i.e. it is an equivalence relation that is additionally compatible with all of the ring operations addition and multiplication, i.e. the congruence satisfies the following Congruence Sum & Product Rules

$$\begin{align} a_1\equiv b_1\\ a_2\equiv b_2\end{align}\ \Rightarrow\ \begin{array}{}a_1 + a_2\equiv b_1+b_2\\ a_1 \times a_2\equiv b_1 \times b_2 \end{array}\qquad$$

This implies that the ring operations descend to well-defined operations on the equivalence classes $\,[a] = a+n\Bbb Z\,$ via $\,[a]+[b] := [a+b],\ [a]\times [b] := [a\times b],\,$ and the map $\, a\mapsto [a]\,$ is a surjective ring hom, which immediately implies that all the ring laws persist in the image $\,\Bbb Z_n\,$, so $\,\Bbb Z_n\,$ has associative and commutative addition and multiplication, connected via the distributive law, so arithmetic in $\,\Bbb Z_n\,$ is essentially the same as in $\,\Bbb Z,\,$ except that some elements have been forced to be equal.

For computational purposes it is often convenient to map the classes to normal (canonical) representatives $\,h\,:\, [a]\mapsto \bar a.\,$ The most common choice is its least nonnegative element $\,\bar a := a\bmod n\,$, but also convenient are least magnitude reps, e.g. $\,0,\pm1,\pm2\pmod{\!5}.\,$ Generally we can use any complete system of residues, i.e. any $\,n\,$ integers such that every integer is congruent to one in our set.

Then we can transport the ring structure to the normal forms by pulling (back) the ring operations along $h$ to obtain the induced ring operations on the normal forms as follows:

$$\bar a + \bar b = h([a])+h([b]) = h([a]+[b]) = h([a+b]) = \overline{a+b}$$

e.g. this becomes $\ a\bmod n + b\bmod n\, = \, (a+b)\bmod n\ $ using the common normal forms.

Said equivalently, to perform an operation on normal forms $\,\bar a,\,\bar b,\,$ we apply $h^{-1}$ to map them to their associated classes $\,[a],[b],\,$ then we perform the operation on the classes yielding $\,[a+b],\,$ then finally we apply $h$ to map that result to its normal form $\,\overline{a+b}.\,$ So the normal forms are essentially "labels" or "names" for their congruence classes. We could instead use any set of $n$ elements as labels, but using a subset of the original ring $\,\Bbb Z\,$ makes more intuitive how the normal form corresponds to the class.

As above, in Euclidean domains like $\,\Bbb Z\,$ and $\,F[x]\,$, which enjoy Euclidean division with smaller remainder, it is convenient to use the the remainder as the normal rep, which is discussed further here, showing how Hamilton's pair representation for complex numbers is just a special case of this (in $\,\Bbb R[x]\bmod x^2\!+\!1\cong \Bbb R[i]\cong \Bbb C)$.