Any Non-trivial Logic System Defined with only Equality

Question

Preface

We can define a system of logic by conjunctions of rules that system must follow. For example, if we wanted to define transitivity of numbers: $$F_1: \forall a,b,c. a < b \land b < c \rightarrow a < c$$ We could similarly define associativity: $$F_2: \forall d,e,f. (d + e) + f = d + (e + f)$$ Then we could define a system that follows these rules by the conjunction of all these rules: $$F : F_1 \land F_2$$ Then if we wanted to check if a statement is satisfiable, say $S: c + 2 = b \land b = 1 \land c = 0$, we could take the conjunction of $S \land F$ and then calculating their congruences classes to see if any contradictions arise.

Question

I am curious: Can any valid non-trivial systems of logic be describe only using equalities ($=$) and conjunctions ($\land$)?

What I mean by non-trivial is that for the logical system in question, $F$, there exists some statements $S$, only using conjunction and equality for both $F$ and $S$, such that: $$SAT(S \land F) \neq SAT(S)$$ Where $SAT$ evaluates to $\top$ or $\bot$ if the logic is satisfiable or not.

This is to say, the conjunction of $F$ with $S$ changes the satisfiability of $S$. If $S$ were already satisfiable, $F$ may define a system where $S$ is in fact a contradiction.

Example 1

As an example say: $$S: x = 1 \land x = 0$$ This is satisfiable by $x = 1 = 0$, because we haven't defined what the relation of 1 and 0 are. If we introduce a system of logic: $$F: 1 \neq 0$$ Then take their conjunction: $$1 \neq 0 \land x = 1 \land x = 0$$ This is no longer satisfiable. You can see this is mainly because we introduced an inequality from $F$. I am wondering if you could define such an $F$ only using equality and conjunction that changes the satisfiability of an $S$.

Example 2

I don't think this is possible. Let's say, for instance we define a system of logic $I$ that defines a set of integers and the addition operation ($+$), conjunctions of associativity, transitivity, reflexivity, axioms, etc. Let's assume we do this only using equality and conjunctions.

Then we describe a statement: $$S: a = b \land a = b + 1$$

$S$ is perfectly satisfiable on it's own because they would all be in the same congruence class, and because $+$ is not defined we have $a = b = b + 1$. Then introduce $I$: $$I \land S$$ Clearly this should not be satisfiable. However, if we have only used equality then there is nothing to let us know that any particular congruence class will have a conflict and thus it is still satisfiable (I think). Thus resulting in a contradiction that we have not actually define the logic system of integers. From this, I would conjecture that this is not possible, as you would need some inequality ($<, \neq, >$) to refute or at least discover a contradiction.