Equality in first-order logic

First-order logic (FOL) can be discussed with and without regarding identity as a in-built relation. This difference matters because the semantic of identity ("anything is equal to nothing but itself") is not definable in FOL. Hence, different ways to express the concept of identity lead to different models. By contrast, identity is definable in SOL, so the issues discussed here is not a problem in SOL.

First-order logic with identity

FOL with identity is a logic where "$=$" is a primitive logic symbol, i.e., symbols with the same semantics across all possible model. As a logical symbol, we stipulate that any model $\mathcal A$ of the logic always interpreted "$=$" to the identity relation on ${\rm Dom}(\mathcal A).$ Furthermore, the incorporation of equality adds three "Axioms of Equality" to the deduction system of the logic: reflexivity and the substitution for formulas, where the second axiom (aka. Leibniz's law) is an axiom schema. Other properties such as symmetry and transitivity can be derived from these two axioms.
Remark. Note that the Axioms of Equality are actually not axioms but inference rules. In fact, these axioms do NOT characterise equality––what they characterise are arbitrary equivalence relations on the domain of discourse that is congruent with respect to the functions and relations of the interpretation.
Example. Peano Arithmetic is the FO theory of $(\omega, +, succ)$ with identity, but Modulo Arithmetic is a FO theory without identity. Instead, for a fixed $n>0$, an equivalence relation $\equiv_n$ is defined such that $a \equiv_n b$ iff $a \equiv b$ mod $n$.

First-order logic without identity

Consider a vocabulary $\sigma$ where equality is not a logical symbol. A theory that contains the Axioms of Equality is called a theory with equality. Let $T$ be such a theory of $\sigma$ where the Axioms of Equality, denoted as $A_{\rm eq}$, are defined for a binary relation $\sim$. Given a model $\mathcal A$ of $T$, it is possible that $a\sim^{\mathcal A} b$ but $a\neq b$ although $a$ and $b$ will satisfy exactly the same formulas in ${\mathcal A}$ by Leibniz's law. On the other hand, we can define the quotient structure $\mathcal B := \mathcal A/\!\!\sim$ of $\mathcal A$. Clearly, $a\sim^{\mathcal B} b$ if and only if $a=b$.

Now we have two methods to define equality in FOL. What would be the difference anyway? The first thing to note is that satisfiability is independent of which method is used. To see this, let $F$ be a formula of vocabulary $\sigma$ plus equality and let $F'$ be the formula obtained by replacing all "$=$" in $F$ with "$\sim$". Consider the theory $T_F := A_{\rm eq}\cup\{F'\}$. Note that if $\mathcal A$ is a model of $T_F$ then $\mathcal A/\!\!\sim$ is a model of $F$. It follows that $F$ and $T_F$ are equi-satisfiable, as concluded in the following facts:
Fact. A formula $F$ is satisfiable in FOL with identity iff $T_F$ is satisfiable in FOL w/o identity.
Fact. A formula $F$ is valid in FOL with identity iff $A_{\rm eq}\implies F$ is valid in FOL w/o identity.
Let $T$ be a theory of a vocabulary with $\sim$. A model $\mathcal A$ is called a normal model of $T$ if $\mathcal A$ is a model of $T$ and $a\sim^{\mathcal A} b$ implies $a=b$. Hence, a theory with equality has a model iff it has a normal model. Results about FOL in the literature often assume identity with the notion of normal model, these results can be easily rephrased without identity. For example,
Theorem. (Compactness with identity) Given a set of formula $\Sigma$, if any finite subset of $\Sigma$ has a model, then $\Sigma$ also has a model.
Theorem. (Compactness w/o identity) Given a set of formula $\Sigma$, if for any finite subset of $\Sigma$ has a normal model, then $\Sigma$ also has a normal model.

References

The Logic of Equality, Leon Henkin, 1977.