It seems that the text is using the lemma (arithmetized $Σ_1$-completeness of PA) for $Σ_1$-formulae rather than just sentences. Originally, I had thought that the generalized version could be easily proven from the specialized one, but I made a careless mistake. Now I believe that it cannot be proven in such a way.
$
\def\pa{\text{PA}}
\def\prov{\text{Prov}}
\def\prf{\text{Proof}}
\def\code#1{\ulcorner#1\urcorner}
\def\num#1{\underline{#1}}
\def\vv{\vec{v}}
$
First I shall give the generalized theorem and an outline of its proof. I shall use the provability modal operator where $⬜φ$ is some sentence that says "$φ$ is provable after its free variables have each been substituted by a numeral encoding its value". For example $⬜( \ ∀x{<}k\ ( \ x·x<k·x \ ) \ )$ expands to $\prov(\code{ ∀x{<}\num{k}\ ( \ x·x<\num{k}·x \ ) })$.
Theorem: Take any $Σ_1$-formula $φ$ with free variables $\vv$. Then $\pa ⊢ ∀\vv\ ( \ φ→⬜φ \ )$.
Proof: (Work with a deductive system for FOL that permits proving formulae with free variables, which are implicitly universally quantified.) Let $ψ$ be a formula equivalent to $φ$ that is in prenex normal form with only bounded universal quantifiers and with matrix in disjunctive normal form. We can assume that every literal in $ψ$ is "$x+y=z$" or "$x·y=z$" for some variables/numerals $x,y,z$, by trichotomy and using $x<y ≡ ∃d\ ( \ x+d+1=y \ )$ and de-nesting function-symbols. (For example, $x·y<z·z$ $≡ ∃a,b,c,d\ ( \ x·y=a ∧ a+1=b ∧ z·z=c ∧ a+d=c \ )$.) Then it suffices to show that $\pa ⊢ ψ→⬜ψ$, because $\pa ⊢ φ→ψ$ and $\pa ⊢ ⬜( \ ψ→φ \ )$. Note that:
(1) $\pa ⊢ x+y=z → ⬜( \ x+y=z \ )$, for any variables/numerals $x,y,z$. [By induction.]
(2) $\pa ⊢ x·y=z → ⬜( \ x·y=z \ )$, for any variables/numerals $x,y,z$. [By induction.]
(3) $\pa ⊢ ⬜α∧⬜β → ⬜( \ α∧β \ )$, for any formulae $α,β$.
(4) $\pa ⊢ ⬜α∨⬜β → ⬜( \ α∨β \ )$, for any formulae $α,β$.
(5) $\pa ⊢ ∃x\ ( \ ⬜α \ ) → ⬜( \ ∃x\ ( \ α \ ) \ )$, for any formula $α$ and variable $x$.
[Because $\pa ⊢ (⬜α)[x{:=}c] → ⬜( \ α[x{:=}c] \ )$.]
(6) $\pa ⊢ ∀x{<}t\ ( \ ⬜α \ ) → ⬜( \ ∀x{<}t\ ( \ α \ ) \ )$, for any formula $α$ and variable $x$ and term $t$.
[By induction with respect to $t$, since $\pa ⊢ ∀x{<}t\ ( \ α \ ) ∧ α[x{:=}t] ↔ ∀x{<}t{+}1\ ( \ α \ )$.]
By induction on the logical structure of $ψ$, using (1) and (2) on the literals in the matrix of $ψ$ and then (3) to (6) repeatedly, we obtain the desired claim.
In case you want a reference for the generalized lemma, I managed to find it in Rautenberg's "A Concise Introduction to Mathematical Logic" in Theorem 2.1 under Section 7.2 on "The Provable $Σ_1$-Completeness". Rautenberg did not clearly indicate disparity between the generalized and the specialized versions, but I feel that there is no easy way to bootstrap, because the induction I used in the above proof has parameters arising from those free variables.
To reiterate the comments, 2 is correct, but 1 is not. The author's argument that $\sf PA\vdash Con^{\circ}$ is essentially the same as your argument that $\sf Con^{\circ}$ is a trivial logical truth and doesn't transfer to $\sf Con^\star.$ And as you say, $\sf Con^\star$ expresses that there is no least proof of inconsistency, so this is provably equivalent to $\sf Con$ in $\sf PA$ as an instance of the least number principle.
So actually, the $\le$ (as opposed to $<$) is crucial. But for that matter, I'm not sure why the author doesn't just use the simpler definition $$\sf Prf^\circ(x,y) := Prf(x,y)\land \lnot Prf(x, \ulcorner \bot \urcorner).$$ Unless I'm missing something, one can show $\emptyset \vdash \sf Con^\circ$, and that $\sf Prf^\circ$ captures and expresses $Prf$ iff PA is consistent, etc. under this simpler definition by similar arguments.
Best Answer
Below I suppress Godel numbering for brevity.
I think bringing in the standard model makes things less clear. The key point is that PA proves its own $\Sigma_1$-completeness, that is, for each $\Sigma_1$ sentence $\varphi$ we have $\mathsf{PA}\vdash\varphi\rightarrow\mathrm{Prov}_\mathsf{PA}(\varphi)$ (that's crucially just a one-way implication of course). Taking contrapositive of the right hand side and using the fact that the negation of a provability statement is a consistency statement, this is just $$\mathsf{PA}\vdash \mathrm{Con}_\mathsf{PA}(\neg\varphi)\rightarrow\neg\varphi,$$ whence by the deduction theorem we get $$\mathsf{PA}+\mathrm{Con}_\mathsf{PA}(\neg\varphi)\vdash\neg\varphi$$ as desired.
Now just replace "$\neg\varphi$" by "$\psi$" - if $\varphi$ is arbitrary $\Sigma_1$ then $\psi$ is arbitrary $\Pi_1$.