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 $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.
Here is an minimal example satisfying both conditions at the same time, maybe not super interesting in itself but there we go.
Let $C$ and $D$ be two subcategories of the category of sets (so in particular they are concrete categories), defined by:
the objects of $C$ are $\{1\}$ and $\{2\}$, and the only morphisms are the identities, and the only map $\{1\}\to \{2\}$ (we do not take the map $\{2\}\to \{1\}$;
the objects of $D$ are $\{1,2\}$ and $\{3,4\}$, and the only morphisms are the identities, and the constant map $\{1,2\}\to \{3,4\}$ with value $4$.
Then those two categories are equivalent (they are both equivalent to the abstract category $\bullet\to \bullet$), but the non-identity map in $C$, which is a bijection, is sent to the map in $D$ which is neither injective nor surjective.
Best Answer
Well, the actual difference between the two statements is that for an equivalence of categories, we only require that that the composites $F \circ G$ and $G \circ F$ are naturally isomorphic to the identity functors rather than exactly equal. That is, there's a collection of isomorphisms $\eta_x :GF(x) \rightarrow x$ for each object of $A$ such that whenever $f: x \rightarrow y$ is a morphism in $A$, $\eta_y GF(f) = f \eta_x$, and a similar natural isomorphism for $F \circ G$.
As for why we do this... Imagine we're both doing group theory, so we both get ourselves a category of groups and start doing group theory in that category. But then we compare our categories and they're not the same: your category has one object for each isomorphism class of groups, while the objects of my category are given by a set $X$ along with a multiplication $\otimes: X \times X \rightarrow X$ which makes it a group.
Our categories aren't isomorphic, not by a long shot: for every object in your category there's a large class of objects in mine. So if we could only use isomorphisms of categories it would look like we're working on entirely different things.
Fortunately, our two categories are equivalent: using one functor which sends a set and a multiplication to its isomorphism class, and the other functor which takes each isomorphism class and picks a realisation of that group. Therefore, we're justified in calling both categories 'The Category of Groups' and any result you get in your category will also work in mine.