Claim. The following are equivalent for a limit ordinal $\alpha$ :
$\alpha$ is a beth-fixed point.
$V_\alpha$ thinks every set is equipotent with an ordinal.
For $1\to 2$, it suffices to show that for every $\beta<\alpha$, $V_\alpha$ thinks $V_\beta$ is equipotent with an ordinal. (This is because every set in $V_\alpha$ is a subset of some $V_\beta$, $\beta<\alpha$.) Let $f:V_\beta\to\beth_\beta$ be a bijection. Then $f$ is a subset of $V_\beta\times \beth_\beta$, which is a member of $V_\alpha$. Since $V_\alpha$ is closed under Cartesian products and power sets, $f\in V_\alpha$.
For $2\to 1$, observe that the assumption implies $|V_\beta|\in V_\alpha$ for every $\beta<\alpha$, so $\beth_\beta<\alpha$ for all $\beta<\alpha$, which means $\alpha$ is a $\beth$-fixed point.
I finally prove that the above characterization is equivalent to the validity of $\Sigma_1$-replacement over $V_\alpha$:
Claim. If $\alpha$ is a beth-fixed point, then $V_\alpha$ satisfies $\Sigma_1$-Replacement.
The main ingredient is the following version of Levy reflection principle (which is provable by the same proof of usual Levy reflection principle $H_\kappa\prec_{\Sigma_1} V$)
Theorem. Let $\lambda<\kappa$ be cardinals and $\lambda$ be regular. Then $H_\lambda\prec_{\Sigma_1} V_\kappa$.
Moreover, it is known that $H_\lambda$ is a model of ZFC without Power set if $\lambda$ is regular.
Now let $F$ be a $\Sigma_1$-class function over $V_\alpha$ with a parameter $p$. Take $x\in V_\alpha$. Choose $\xi<\alpha$ such that $p,x\in V_\xi$. Since $\alpha$ is a beth-fixed point, $\lambda:=|V_\xi|^+<\alpha$. We can see that $V_\xi\subseteq H_\lambda\subseteq V_\alpha$.
Observe that $F$ is absolute between $V_\alpha$ and $H_\lambda$. Moreover, $H_\lambda$ satisfies Replacement for $F$. Let $H_\lambda\models F^"[x]=y$ for $y\in H_\lambda$. Since the formula
$$[\forall v\in y\exists u\in x (F(u)=v)]\land [\forall u\in x\exists v\in y (F(u)=v)]$$
is $\Sigma_1$-formula, it also holds over $V_\alpha$. This shows $y$ witnesses the instance of replacement for $F$, $x$ and $p$.
The existence of a proper class which we can prove each relativized ZF axiom does not prove the consistency of ZF. To see the absurdity of this, observe that $V$ is a proper class for which we can prove each relativized axiom... since each relativized axiom is just an axiom we take. So if that sufficed to prove consistency of ZF, we would be in violation of Godel's theorem from the get-go.
In order to prove the consistency of ZF within ZF, we would need to find a model that we could internally prove that every axiom holds. In other words we can prove the internally quantified statement "for every axiom $\ulcorner\phi\urcorner$ of ZF, $M\models \ulcorner\phi\urcorner$." (I use "Quine corners" to emphasize this is an internal code for the formula as a set.) This is different from the sense in this theorem, which is to say for every axiom $\phi$ of ZF we can prove "$\phi^M$."
Best Answer
No, the compactness theorem plus a result specific to $\mathsf{ZFC}$ prevents this. Suppose $T$ were such a set of sentences. Then the union of $T$ and the replacement scheme is unsatisfiable, so by compactness there must be some set $F$ of finitely many replacement instances such that $T\cup F$ is inconsistent. But this means that, over $\mathsf{ZC}$ (= set theory without replacement), $F$ implies the whole replacement scheme. However, this is known not to be possible; in fact, $\mathsf{ZFC}$ proves the consistency of every extension of $\mathsf{ZC}$ by finitely many replacement instances, so $\mathsf{ZFC}$ is "very far" (via Godel's second incompleteness theorem) from any finite extension of $\mathsf{ZC}$.