If you are allowed to use soundness and completeness theorems (which hold not only in propositional logic but also in first-order logic), the proof is quite easy.
$\Leftarrow$: Suppose $\Sigma \cup \{\neg\varphi\}$ is inconsistent, then there exists a formula $\psi$ such that $\Sigma \cup \{\neg\varphi\}\vdash\psi$ and $\Sigma\,\cup \{\neg\varphi\}\vdash\neg\psi$.
By soundness theorem, $\Sigma \cup \{\neg\varphi\} \models \psi$ and $\Sigma \cup \{\neg\varphi\}\models \neg\psi$, which means that every model of $\Sigma \cup \{\neg\varphi\}$ satisfies both $\psi$ and $\lnot \psi$. Now, by definition, there is no structure in first-order logic that satisfies a formula and its negation.
Therefore, there is no model of $\Sigma \cup \{\neg\varphi\}$, which means that every model of $\Sigma$ is a model of $\varphi$ as well, i.e. $\Sigma \models \varphi$.
According to completeness theorem, $\Sigma \vdash \varphi$.
$\Rightarrow:$ Suppose $\Sigma \vdash \varphi$. Then, $\Sigma \cup \{\lnot \varphi\} \vdash \varphi$, according to the weakening property (which holds in any deduction system for "traditional logics").
But $\Sigma \cup \{\neg\varphi\} \vdash \lnot \varphi$ as well, since $\lnot \varphi \in \Sigma \cup \{\neg\varphi\}$. Hence, $\Sigma \cup \{\neg\varphi\}$ is inconsistent. $\qquad\square$
Note the your use of the soundness theorem in your original post is not correct, or at least not well-written: according to soundness theorem, $\Sigma \vdash \varphi$ does not imply that $\varphi$ is logically valid (a formula is logically valid iff every structure satisfies it), but it implies that every model of $\Sigma$ satisfies $\varphi$ (it is possible that $\Sigma$ has no models and $\varphi$ is unsatisfiable as well).
Roughly, weakening property says that if you can prove something starting from some hypothesis $\Sigma$, you can prove it even when you add more hypotheses to $\Sigma$.
If you are not allowed to use soundness and correctness theorems, the proof of ($\Leftarrow$) is slightly more technical and it depends on the deduction system and the inference rules you are allowed to use. This means that soundness and completeness theorems are not necessary to prove ($\Leftarrow$) but they simplify the proof and also they "universalize" the proof, in the sense that by appealing to these theorems the proof of ($\Leftarrow$) does not depend explicitly on the deduction system and the inference rules you are allowed to use.
Anyway, if the deduction system you are using is Hilbert system with the axiom $(\lnot \varphi \to \lnot \psi) \to ((\lnot \varphi \to \psi) \to \varphi)$ (as the systems described in Mendelson's book, pp. 35 and 69 for propositional and first order logic, respectively), you can easily prove $(\Leftarrow)$ as described in this post, without appealing to the semantic notions involved in soundess and completeness theorems.
Usually classical first-order logic is defined so that classical propositional logic is a literal sublanguage of it. In particular, it is the sublanguage where all relation symbols are nullary (and omitting quantifiers which are essentially useless at that point). This makes it impossible for there to be any place to put terms. Semantically, it then doesn't matter what the domain is as there is no way to refer to any of its elements. The tautological implication relation is the logical implication relation restricted to this sublanguage.
Nevertheless, what you describe works insofar as the theorem in your last bullet holds, but it is kind of a weird way to go about it. You can improve and simplify it by making it a first-order theory rather than constraining the interpretation in an ad-hoc manner. You could add the axioms $\exists x.P(x)$ and $\exists x.\neg P(x)$ which would ensure the domain contains at least two elements and the interpretation of $P$ is non-empty and has a non-empty complement. It may be a bit clearer to just explicitly add two constants $\mathsf{tt}$ and $\mathsf{ff}$ and the axioms $P(\mathsf{tt})$, $\neg P(\mathsf{ff})$, and $\mathsf{tt}\neq\mathsf{ff}$.
There are many other ways to map propositional formulas to FOL formulas such that the original formulas are tautologies if and only if the resulting formula is valid. One notable one is to simply formalize the notion of a Boolean algebra. Alternatively, if you went the route you did because Enderton doesn't allow for nullary relation symbols (or maybe it just didn't occur to you), you could do essentially the same thing I suggested but producing distinct unary predicates for each propositional variable (sentence symbol). Then, to get effectively nullary predicates you could map the propositional variables to formulas like $\forall x.P_n(x)$ (or instead $\exists x.P_n(x)$ or $P_n(\mathsf{c})$ after adding a constant $\mathsf{c}$). This is clearly a bit of a hack to make up for not directly having nullary relation symbols.
Best Answer
As someone pointed out to me somewhere else, my mistake is not noting the difference between "$\Gamma$ tautologically implies $\varphi$" and "$\Gamma\cup\Lambda$ tautologically implies $\varphi$". It is indeed the case that if $\varphi$ is logically implied by $\Gamma$ then $\Gamma\cup\Lambda$ tautologically imply $\varphi$; it does not mean that $\Gamma$ alone tautologically implies $\varphi$.