= Solution
An <algebra involution> is a map $x\mapsto x^*$ satisfying
$$
(\alpha x+\beta y)^*=\overline\alpha x^*+\overline\beta y^*,\qquad
(xy)^*=y^*x^*,\qquad (x^*)^*=x.
$$
Thus it is a <conjugate-linear map>, reverses multiplication, and has square equal to the identity map. These are algebraic axioms; <norm> <continuity>, when included in a convention for a Banach star algebra, is an additional requirement and is not used here. A <Hermitian element of a star algebra> satisfies $h^*=h$.
Conjugate <linearity> gives $0^*=0$. The map $*$ is surjective, so every $u$ is $v^*$ for some $v$. The identities $(v1)^*=1^*v^*=v^*$ and $(1v)^*=v^*1^*=v^*$ show that $1^*$ is a two-sided identity; uniqueness of the identity gives $1^*=1$. Therefore \b[both zero and the identity are <Hermitian algebra elements>].
For the final unheaded request, first prove equality of the two spectra for a <Hermitian element of a star algebra> $h\in B$. Write $D=\mathbb C\setminus\sigma_A(h)$. The spectrum is a <compact> subset of the real axis by the assumption that $A$ is a <Hermitian Banach algebra>, so $D$ is <connected>: points above and below the real axis can be joined by paths passing beyond the endpoints of an interval containing the spectrum; points of the real axis outside the spectrum can first move a short distance vertically. Define
$$
E=\{\lambda\in D:(\lambda1-h)^{-1}\in B\}.
$$
For large $|\lambda|$, the <Neumann series> $\lambda^{-1}\sum_{n\ge0}(h/\lambda)^n$ converges in the <closed> unital <star-subalgebra> $B$, so $E$ is nonempty. It is relatively open in $D$ by the same series applied to a perturbation of an already invertible element of $B$. It is relatively <closed> in $D$, since the <resolvent of an element> is <continuous> in $A$ and $B$ is <closed>. Connectedness gives $E=D$. Conversely an inverse in $B$ is an inverse in $A$, so
$$
\sigma_B(h)=\sigma_A(h).
$$
This proves the needed <spectrum in a closed unital subalgebra> equality for <Hermitian algebra elements> without presupposing that their spectra in $B$ are real.
Now suppose $b\in B$ is invertible in $A$. Both $bb^*$ and $b^*b$ are <Hermitian algebra elements>, since reversing products and applying $*$ twice fixes them. They are invertible in $A$, by part (ii), so the preceding equality of spectra of <Hermitian algebra elements> makes them invertible in $B$. Part (ii), applied within $B$, then makes $b$ invertible in $B$. Applying this to $b-\lambda1$ for every scalar $\lambda$, and using the automatic converse for an inverse already in $B$, proves <spectral permanence for Hermitian Banach algebras>:
$$
\boxed{\sigma_B(b)=\sigma_A(b)\quad(b\in B).}
$$
Only closedness, a common identity, closure under the <algebra involution>, and real ambient spectra of <Hermitian algebra elements> were used; the <C-star identity> was not assumed.
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