Well, I decided to include some additional information.
Definition 1. Let $\mathcal{C}$ and $D$ be categories, $T\colon\mathcal{C}\to\mathcal{D}$ and $S\colon\mathcal{D}\to\mathcal{C}$ be functors. Then the pair $(T,S)$ is called an equivalence iff $S\circ T\cong I_{\mathcal{C}}$ and $T\circ S\cong I_{\mathcal{D}}$. In this case functors $T$ and $S$ are also called equivalences, and categories $\mathcal{C}$ and $\mathcal{D}$ are called equivalent.
It is a basic definition and Tim's answer shows why we need to use equivalence even more frequently than isomorphism. Here's another important definition:
Definition 2. Let $\mathcal{C}$ be a category, $\mathcal{S}$ be a subcategory of $\mathcal{C}$. Then the category $\mathcal{S}$ is called a skeleton of $\mathcal{C}$ iff it is a full subcategory of $\mathcal{C}$ and every object of $\mathcal{C}$ is isomorphic to precisely one object of $\mathcal{S}$.
Note, that if the axiom of choice holds, then every category has a skeleton. See also nLab article. The connection between equivalences of categories and their skeletons is described in the following proposition:
Proposition 1. Let $\mathcal{C}$ and $\mathcal{D}$ be categories, $\mathcal{S}_{\mathcal{C}}$ and $\mathcal{S}_{\mathcal{D}}$ be their skeletons. Then $\mathcal{C}\simeq \mathcal{D}$ iff $\mathcal{S}_{\mathcal{C}}\cong\mathcal{S}_{\mathcal{D}}$.
The proof follows from the fact that every category is equivalent to its skeleton and if two skeletal categories are equivalent, then they are isomorphic. You can also search Mac Lane's "Categories for the working mathematician" for the details.
Thus we can use the notion of skeleton instead of the original definition of equivalence, but sometimes it is not a simplification. As it was mentioned, even an attempt to prove that a category has a skeleton may lead to the set-theoretical difficulties. Tim also gave arguments.
You write: But why go through this whole process of defining this notion of equivalent categories if we can just create a single equivalence classes of objects via the equivalence relation of being isomorphic, and make morphisms defined on the same equivalence class "the same?"
Okay, it could be a good idea if we want to define something like a skeleton. But it isn't, because the straightforward applying this idea leads to wrong definition. Let's try to do this.
Definition 3. Let $\mathcal{C}$ be a category. Then define the graph $\text{Equiv}(\mathcal{C})$ in the following way: $\text{Obj}(\text{Equiv}(C))=\text{Obj}(\mathcal{C})/\cong_{\mathcal{C}}$ and $$\text{hom}_{\text{Equiv}(\mathcal{C})}([a],[b])=(\coprod_{a'\in[a],b'\in[b]}\text{hom}_{\mathcal{C}}(a',b'))/[(f\sim g)\Leftrightarrow(\exists a,b\in \text{Iso}(\mathcal{C})|\quad g\circ a=b\circ f) ].$$
But the graph $\text{Equiv}(\mathcal{C})$ doesn't inherit the composition law from $\mathcal{C}$. The graph $\text{Equiv}(\mathcal{C})$ doesn't even coincide with graph of any skeleton of $\mathcal{C}$ in general case. For example, it may paste two morphisms with the same domain: in the category $\mathbf{Finord}$ we have $\text{end}_{\text{Equiv}(\mathbf{Finord})}([2])=\text{hom}_{\text{Equiv}(\mathbf{Finord})}([2],[2])\cong2$, but $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.
A standard way of trying to explain how an isomorphism between two objects of any category means precisely that they look exactly the same as far as that category is concerned is via the Yoneda embedding.
In our situation we should observe that two objects $x,y$ in a category $C$ are isomorphic if and only if they represent the same functor, that is, if and only if there is a natural isomorphism between the functors $Hom_C(-,x)$ and $Hom_C(-,y)$. In other words, $x$ and $y$ are isomorphic if and only if they admit isomorphic sets of maps from all other objects of $C$. Since a natural isomorphism is merely a natural transformation whose components are bijections of sets, this actually reduces the general notion of isomorphism to the algebraic concept of "bijective morphism", showing that in this sense every category is a category of algebraic structures. In particular, this is not circular-we don't already need to know what an isomorphism is in general to know what a natural isomorphism is!
So, if we define "$x$ and $y$ have precisely the same structure" as "$x$ and $y$ look the same as codomains to all other objects of $C$", formally, $x$ and $y$ represent naturally isomorphic functors, then the conservativity of the Yoneda embedding proves that our notion of isomorphism is the correct one.
One possible complaint is that you might wonder why we shouldn't say that $x$ and $y$ are "the same" if they admit the same maps to every other object of $C$, rather than from. Luckily, these are equivalent, since objects are isomorphic in $C$ if and only if they are isomorphic in $C^\mathrm{op}$. The main remaining complaint is that we might somehow want more of $x$ and $y$ than that the other objects of $C$ see them as being the same. But in that case, it seems clear to me that we're no longer talking about any notion of sameness in $C$.
Best Answer
Consider the category of sets and the operation of the cartesian product of sets. It is rather straightforward to prove that the cartesian product operation is associative up to canonical isomorphisms. This is simply a matter of moving parenthesis in the syntactic description of the cartesian products as tuples but also follows from the universal property of products. Now, you define the notion of cardinality of a set as follows. Let $C$ be the category of sets and bijections and let $Card$ be a skeleton of it. There is an inclusion $Card \to C$ which is an equivalence but not an isomorphism. Now we wish to define, say, the product of cardinalities. We do so by $Card \times Card \to C\times C \to C \to Card$ using the injection followed by the cartesian product followed by the quasi-inverse of the inclusion. Now we phrase the problem: is the product of cardinalities associative? To answer that we translate the problem back to $C$ where we do not get an equality between the sets representing the problem but we do get an isomorphism. That isomorphism tells us that back in $Card$ we do have an equality. So the solutions in $C$ do not match exactly with the solutions in $Card$. But the equivalence is sufficient.