To amplify, let me point out that it is not even true that an additive functor that preserves epimorphisms/surjections also preserves cokernels. For example, let $\mathcal{C}$ be the full subcategory of $\textbf{Ab}$ spanned by the torsion-free abelian groups. This category, perhaps unexpectedly, is an additive category with kernels and cokernels – but is not an abelian category. Indeed, in $\mathcal{C}$, the cokernel of $2 \cdot {-} : \mathbb{Z} \to \mathbb{Z}$ is $0$, so the inclusion $\mathcal{C} \hookrightarrow \textbf{Ab}$ is an additive functor that preserves surjections but not cokernels (or even epimorphisms in general!). In fact, $\mathcal{C} \hookrightarrow \textbf{Ab}$ is even a right adjoint, and so is left exact in particular. (When I say left/right exact, I always mean a functor that preserves finite limits/colimits.)
However, what is true is that a left exact functor $F : \mathcal{A} \to \mathcal{B}$ that preserves (normal) epimorphisms will be exact, provided $\mathcal{A}$ and $\mathcal{B}$ are both abelian. Indeed, since $\mathcal{A}$ and $\mathcal{B}$ are additive, to show that $F$ is right exact it is enough to show that it is additive and preserves cokernels. Now, it is known that a functor $\mathcal{A} \to \mathcal{B}$ that preserves finite products is automatically additive; but a left exact functor preserves finite products, so is an additive functor in particular. Consider a sequence of morphisms
$$0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0$$
in $\mathcal{A}$, and suppose $A' \to A$ is kernel of $A \to A''$ and $A \to A''$ is the cokernel of $A' \to A$. Since $F$ preserves kernels, we get an exact sequence
$$0 \longrightarrow F A' \longrightarrow F A \longrightarrow F A''$$
in $\mathcal{B}$, and since $A \to A''$ is a (normal) epimorphism, we can extend the above to a short exact sequence in $\mathcal{B}$:
$$0 \longrightarrow F A' \longrightarrow F A \longrightarrow F A'' \longrightarrow 0$$
Thus, $F$ preserves cokernels of normal monomorphisms. In general, if we have a morphism $X \to A$ in $\mathcal{A}$, we can factor it as $X \to A' \to A$ where $X \to A'$ is the cokernel of the kernel of $X \to A$, and it is not hard to show that the cokernel of $A' \to A$ is also the cokernel of $X \to A$. Since $F$ preserves all kernels and also cokernels of (normal) monomorphisms, $F$ preserves this factorisation, and therefore the cokernel of $F A' \to F A$ is also the cokernel of $F X \to F A$. However, because $\mathcal{A}$ is an abelian category, $A' \to A$ itself is a (normal) monomorphism, so $F$ preserves its cokernel. Thus $F$ actually preserves all cokernels and is therefore right exact.
Dually, of course, a right exact functor between abelian categories is exact if and only if it preserves (normal) monomorphisms. This explains the classical fact that $M$ is flat if and only if $- \otimes_R M$ preserves injective homomorphisms: $- \otimes_R M$ is a left adjoint and so right exact in particular.
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 $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.
Best Answer
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.