The two claims are true.
Proof for the first claim :
First of all, let us prove by induction on $i$ that
$$S_i=2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})\tag1$$
where $s:=c-\sqrt{c^2-4},t:=c+\sqrt{c^2-4}$.
For $i=0$, $(1)$ holds since
$$S_0 =P_{a}(c)=2^{-a}(s^a+t^a)$$
Supposing that $(1)$ holds for $i$ gives
$$\begin{align}S_{i+1}&=P_a(S_i)
\\\\&=2^{-a}\bigg(\bigg(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})-\sqrt{(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}}))^2-4}\bigg)^a
\\&\qquad\qquad +\bigg(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})+\sqrt{(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}}))^2-4}\bigg)^a\bigg)
\\\\&=2^{-a}\bigg(\bigg(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})-\sqrt{(2^{-a^{i+1}}(t^{a^{i+1}}-s^{a^{i+1}}))^2}\bigg)^a
\\&\qquad\qquad +\bigg(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})+\sqrt{(2^{-a^{i+1}}(t^{a^{i+1}}-s^{a^{i+1}}))^2}\bigg)^a\bigg)
\\\\&=2^{-a}\bigg(\bigg(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})-2^{-a^{i+1}}(t^{a^{i+1}}-s^{a^{i+1}})\bigg)^a
\\&\qquad\qquad +\bigg(2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})+2^{-a^{i+1}}(t^{a^{i+1}}-s^{a^{i+1}})\bigg)^a\bigg)
\\\\&=2^{-a^{i+2}}(s^{a^{i+2}}+t^{a^{i+2}})\qquad\square\end{align}$$
Let $N:=M_p(a)=\frac{a^p-1}{a-1}$. Then, from $(1)$, we get, using $2(c\pm\sqrt{c^2-4})=(\sqrt{c+2}\pm\sqrt{c-2})^2$,
$$\begin{align}S_{p-1}&=2^{-(a-1)N-1}(s^{(a-1)N+1}+t^{(a-1)N+1})
\\\\&=2^{-(a-1)N-1}(s\cdot s^{(a-1)N}+t\cdot t^{(a-1)N})
\\\\&=2^{-2(a-1)N-1}\bigg(s((\sqrt{c+2}-\sqrt{c-2})^{N})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t((\sqrt{c+2}+\sqrt{c-2})^{N})^{2(a-1)}\bigg)\tag2\end{align}$$
Here, we have, by the binomial theorem,
$$\begin{align}&(\sqrt{c+2}\pm\sqrt{c-2})^{N}
\\\\&=\sum_{k=0}^{N}\binom Nk(\sqrt{c+2})^{N-k}(\pm\sqrt{c-2})^k
\\\\&=\sqrt{c+2}\sum_{j=0}^{(N-1)/2}\binom{N}{2j}(c+2)^{(N-2j-1)/2}(c-2)^{j}
\\&\qquad\qquad\qquad\qquad \pm\sqrt{c-2}\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(c+2)^{(N-2j+1)/2}(c-2)^{j-1}\end{align}$$
So, there are integers $C,D$ such that
$$(\sqrt{c+2}\pm\sqrt{c-2})^{N}=C\sqrt{c+2}\pm D\sqrt{c-2}\tag3$$
where
$$C=\sum_{j=0}^{(N-1)/2}\binom{N}{2j}(c+2)^{(N-2j-1)/2}(c-2)^{j}\equiv \left(\frac{c+2}{N}\right)\equiv 1\pmod N\tag4$$
and
$$D=\sum_{j=1}^{(N+1)/2}\binom{N}{2j-1}(c+2)^{(N-2j+1)/2}(c-2)^{j-1}\equiv \left(\frac{c-2}{N}\right)\equiv 1\pmod N\tag5$$
It follows from $(2)(3)(4)(5)$ and $2^{N-1}\equiv 1\pmod N$ that
$$\begin{align}2^{(2a-2)(N-1)}\cdot S_{p-1}&=2^{-2a+1}\bigg(s((\sqrt{c+2}-\sqrt{c-2})^{N})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t((\sqrt{c+2}+\sqrt{c-2})^{N})^{2(a-1)}\bigg)
\\\\&=2^{-2a+1}\bigg(s(C\sqrt{c+2}- D\sqrt{c-2})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t(C\sqrt{c+2}+ D\sqrt{c-2})^{2(a-1)}\bigg)
\\\\&\equiv 2^{-2a+1}\bigg(s(\sqrt{c+2}- \sqrt{c-2})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t(\sqrt{c+2}+\sqrt{c-2})^{2(a-1)}\bigg)\pmod N
\\\\&\equiv 2^{-a}(s^a+t^a)\pmod N
\\\\&\equiv P_{a}(c)\pmod{M_p(a)}
\end{align}$$
from which
$$S_{p-1}\equiv P_a(c)\pmod{M_p(a)}$$follows.$\quad\blacksquare$
Proof for the second claim :
Let $N:=N_p(b)= \frac{b^p+1}{b+1}$. Then, from $(1)$, we get, similarly as above,
$$\begin{align}2^{2(b+1)(N-1)}\cdot S_{p-1}&=2^{(b+1)N-2b-1}(s^{(b+1)N-1}+t^{(b+1)N-1})
\\\\&=2^{-2b-3}\bigg(t\cdot ((\sqrt{c+2}-\sqrt{c-2})^N)^{2(b+1)}
\\&\qquad\qquad\qquad\qquad +s\cdot ((\sqrt{c+2}+\sqrt{c-2})^N)^{2(b+1)}\bigg)
\\\\&\equiv 2^{-2b-3}\bigg(t\cdot (\sqrt{c+2}-\sqrt{c-2})^{2(b+1)}
\\&\qquad\qquad\qquad\qquad +s\cdot (\sqrt{c+2}+\sqrt{c-2})^{2(b+1)}\bigg)\pmod N
\\\\&\equiv 2^{-2b}\bigg((\sqrt{c+2}-\sqrt{c-2})^{2b}+(\sqrt{c+2}+\sqrt{c-2})^{2b}\bigg)\pmod N
\\\\&\equiv 2^{-b}(s^{b}+t^{b})\pmod N
\\\\&\equiv P_b(c)\pmod N\end{align}$$
from which
$$S_{p-1}\equiv P_b(c)\pmod{N_p(b)}$$
follows.$\quad\blacksquare$
The claim is true.
Proof :
Since the roots of $x^3-2x^2+3x-2=0$ are $x=1,\frac{1\pm i\sqrt 7}{2}$, there exist $A,B,C$ such that
$$S_k=A\bigg(\frac{1-i\sqrt 7}{2}\bigg)^k+B\bigg(\frac{1+i\sqrt 7}{2}\bigg)^k+C\cdot 1^k$$
Using $S_0=0,S_1=S_2=1$, we get $A=\frac{i}{\sqrt 7},
B=\frac{-i}{\sqrt 7}$ and $C=0$ to have
$$S_k=\frac{i}{\sqrt 7}\bigg(\frac{1-i\sqrt 7}{2}\bigg)^k-\frac{i}{\sqrt 7}\bigg(\frac{1+i\sqrt 7}{2}\bigg)^k\tag1$$
Case 1 : $n=7$
Since $\left(\frac{-28}{7}\right)=0$, we get
$$S_{\delta(7)}=S_7=7\equiv 0\pmod 7$$
Case 2 : $n$ is a prime satisfying either $n\equiv 1,2$ or $4\pmod 7$
Since $\left(\frac n7\right)=1$, we get
$$\bigg(\frac{-28}{n}\bigg)=\bigg(\frac{-1}{n}\bigg)\bigg(\frac{2^2}{n}\bigg)\bigg(\frac 7n\bigg)=(-1)^{(n-1)/2}\cdot 1\cdot \frac{(-1)^{3(n-1)/2}}{\left(\frac n7\right)}=1$$So, we have, using $(1)$,
$$\begin{align}2^{n+2}S_{\delta(n)}&=2^{n+2}S_{n-1}
\\\\&=\frac{i}{\sqrt 7}\bigg((1+i\sqrt 7)\left(1-i\sqrt{7}\right)^{n}-(1-i\sqrt 7)\left(1+i\sqrt{7}\right)^{n}\bigg)
\\\\&=-\frac{i}{\sqrt 7}\bigg((1+i\sqrt 7)^n-(1-i\sqrt 7)^n\bigg)-\bigg((1+i\sqrt 7)^n+(1-i\sqrt 7)^n\bigg)
\\\\&=-\frac{i}{\sqrt 7}\sum_{j=0}^{n}\binom nj\bigg((i\sqrt 7)^j-(-i\sqrt 7)^j\bigg)-\sum_{j=0}^{n}\binom nj\bigg((i\sqrt 7)^j+(-i\sqrt 7)^j\bigg)
\\\\&=\sum_{k=1}^{(n+1)/2}\binom n{2k-1}2(-1)^{k+1}7^{k-1}+\sum_{k=0}^{(n-1)/2}\binom n{2k}2(-1)^{k+1}7^k\end{align}$$
Since $\binom{n}{m}\equiv 0\pmod n$ for $1\le m\le n-1$, we have
$$\begin{align}2^{n+2}S_{\delta(n)}&\equiv 2(-1)^{(n+3)/2}7^{(n-1)/2}-2\pmod n
\\\\&\equiv 2(-1)^{(n+3)/2}\frac{(-1)^{3(n-1)/2}}{\left(\frac n7\right)}-2\pmod n
\\\\&\equiv 2(-1)^{2n}-2\pmod n
\\\\&\equiv 0\pmod n\end{align}$$from which we get
$$2^{n+2}S_{\delta(n)}\equiv 0\pmod n$$It follows from $\gcd(2^{n+2},n)=1$ that
$$S_{\delta(n)}\equiv 0\pmod n$$
Case 3 : $n$ is a prime satisfying either $n\equiv 3,5$ or $6\pmod 7$
Since $\left(\frac n7\right)=-1$, we get
$$\bigg(\frac{-28}{n}\bigg)=\bigg(\frac{-1}{n}\bigg)\bigg(\frac{2^2}{n}\bigg)\bigg(\frac 7n\bigg)=(-1)^{(n-1)/2}\cdot 1\cdot \frac{(-1)^{3(n-1)/2}}{\left(\frac n7\right)}=-1$$So, we have, using $(1)$,
$$\begin{align}2^{n+1}S_{\delta(n)}&=2^{n+1}S_{n+1}
\\\\&=\frac{i}{\sqrt 7} \bigg((1-i\sqrt 7)\left(1-i\sqrt{7}\right)^{n}-(1+i\sqrt 7)\left(1+i\sqrt{7}\right)^{n}\bigg)
\\\\&=-\frac{i}{\sqrt 7}\bigg((1+i\sqrt 7)^n-(1-i\sqrt 7)^n\bigg)+\bigg((1+i\sqrt 7)^n+(1-i\sqrt 7)^n\bigg)
\\\\&=-\frac{i}{\sqrt 7}\sum_{j=0}^{n}\binom nj\bigg((i\sqrt 7)^j-(-i\sqrt 7)^j\bigg)+\sum_{j=0}^{n}\binom nj\bigg((i\sqrt 7)^j+(-i\sqrt 7)^j\bigg)
\\\\&=\sum_{k=1}^{(n+1)/2}\binom{n}{2k-1}2(-1)^{k+1}7^{k-1}+\sum_{k=0}^{(n-1)/2}\binom{n}{2k}2(-1)^{k}7^k
\\\\&\equiv 2(-1)^{(n+3)/2}7^{(n-1)/2}+2\pmod n
\\\\&\equiv 2(-1)^{(n+3)/2}\frac{(-1)^{3(n-1)/2}}{\left(\frac n7\right)}+2\pmod n
\\\\&\equiv 2(-1)^{2n+1}+2\pmod n
\\\\&\equiv 0\pmod n\end{align}$$from which we get
$$2^{n+1}S_{\delta(n)}\equiv 0\pmod n$$It follows from $\gcd(2^{n+1},n)=1$ that
$$S_{\delta(n)}\equiv 0\pmod n$$
From the three cases above, the claim follows.$\quad\blacksquare$
Best Answer
The claim is true.
Proof :
Let $d:=P^2-6P+1$.
Since the roots of $x^3-Px^2+(2P-1)x-P=0$ are $x=1,\frac{P-1\pm\sqrt{d}}{2}$, there exist $A,B,C$ such that $$S_k=A\bigg(\frac{P-1-\sqrt{d}}{2}\bigg)^k+B\bigg(\frac{P-1+\sqrt{d}}{2}\bigg)^k+C\cdot 1^k$$ Using $S_0=0,S_1=1$ and $S_2=P-1$, we get $A = \frac{-1}{\sqrt{d}}, B =\frac{1}{\sqrt{d}}$ and $C=0$ to have $$ S_k=\frac{-1}{\sqrt{d}}\bigg(\frac{P-1-\sqrt{d}}{2}\bigg)^k+\frac{1}{\sqrt{d}}\bigg(\frac{P-1+\sqrt{d}}{2}\bigg)^k\tag1$$
Also, we get
$$\begin{align}D&=P^2(2P-1)^2-4(2P-1)^3-4P^4-27P^2+18P^2(2P-1) \\\\&=4d\end{align}$$ so $$\bigg(\frac Dn\bigg)=\bigg(\frac{2^2}{n}\bigg)\bigg(\frac{d}{n}\bigg)=\bigg(\frac{d}{n}\bigg)$$
Using $(1)$, we get $$\begin{align}2^nS_{\delta(n)}&=2^nS_n \\\\&=\frac{1}{\sqrt d}\bigg(\bigg(P-1+\sqrt{d}\bigg)^n-\bigg(P-1-\sqrt{d}\bigg)^n\bigg) \\\\&=\frac{1}{\sqrt d}\sum_{j=0}^{n}\binom nj(P-1)^{n-j}\bigg((\sqrt d)^j-(-\sqrt d)^j\bigg) \\\\&=\sum_{k=1}^{(n+1)/2}\binom{n}{2k-1}2d^{k-1}\end{align}$$ Since $\binom nm\equiv 0\pmod n$ for $1\le m\le n-1$, we get $$\begin{align}2^nS_{\delta(n)}&\equiv 2d^{(n-1)/2}\pmod n \\\\&\equiv 0\pmod n\end{align}$$It follows from $\gcd(2^n,n)=1$ that $$S_{\delta(n)}\equiv 0\pmod n$$
Case 2 : $n$ is a prime such that $\left(\frac{d}{n}\right)=1$
Using $(1)$, we get $$S_{\delta(n)}=S_{n-1}=\frac{-1}{\sqrt{d}}\bigg(\frac{P-1-\sqrt{d}}{2}\bigg)^{n-1}+\frac{1}{\sqrt{d}}\bigg(\frac{P-1+\sqrt{d}}{2}\bigg)^{n-1}$$Multiplying it by $2^{n-1}(P-1-\sqrt{d})(P-1+\sqrt{d})=2^{n+1}P$ gives$$\begin{align}&2^{n+1}PS_{\delta(n)} \\\\&=\bigg(-\frac{P-1}{\sqrt{d}}-1\bigg)\bigg(P-1-\sqrt{d}\bigg)^{n}+\bigg(\frac{P-1}{\sqrt{d}}-1\bigg)\bigg(P-1+\sqrt{d}\bigg)^{n} \\\\&=\frac{P-1}{\sqrt{d}}\bigg(\bigg(P-1+\sqrt{d}\bigg)^{n}-\bigg(P-1-\sqrt{d}\bigg)^{n}\bigg) \\&\qquad\qquad-\bigg(\bigg(P-1+\sqrt{d}\bigg)^{n}+\bigg(P-1-\sqrt{d}\bigg)^{n}\bigg) \\\\&=\frac{P-1}{\sqrt{d}}\sum_{j=0}^{n}\binom nj(P-1)^{n-j}\bigg((\sqrt{d})^j-(-\sqrt{d})^j\bigg) \\&\qquad\qquad-\sum_{j=0}^{n}\binom nj(P-1)^{n-j}\bigg((\sqrt{d})^j+(-\sqrt{d})^j\bigg) \\\\&=\sum_{k=1}^{(n+1)/2}\binom n{2k-1}(P-1)^{n-2k+2}2d^{k-1}-\sum_{k=0}^{(n-1)/2}\binom n{2k}(P-1)^{n-2k}2d^{k} \\\\&\equiv 2(P-1)d^{(n-1)/2}-2(P-1)^{n}\pmod n \\\\&\equiv 2(P-1)\bigg(1-(P-1)^{n-1}\bigg)\pmod n \\\\&\equiv 0\pmod n\end{align}$$ since if $P\not\equiv 1\pmod n$, then it follows from $\gcd(P-1,n)=1$ that $(P-1)^{n-1}\equiv 1\pmod n$.
Since we have $2^{n+1}PS_{\delta(n)}\equiv 0\pmod n$, it follows from $\gcd(2^{n+1}P,n)=1$ that $$S_{\delta(n)}\equiv 0\pmod n$$
Case 3 : $n$ is a prime such that $\left(\frac{d}{n}\right)=-1$
Using $(1)$, we have $$\begin{align}&2^{n+1}S_{\delta(n)}=2^{n+1}S_{n+1} \\\\&=\bigg(-\frac{P-1}{\sqrt{d}}+1\bigg)\bigg(P-1-\sqrt{d}\bigg)^{n}+\bigg(\frac{P-1}{\sqrt{d}}+1\bigg)\bigg(P-1+\sqrt{d}\bigg)^{n} \\\\&=\frac{P-1}{\sqrt{d}}\bigg(\bigg(P-1+\sqrt{d}\bigg)^{n}-\bigg(P-1-\sqrt{d}\bigg)^{n}\bigg) \\&\qquad\qquad+\bigg(\bigg(P-1+\sqrt{d}\bigg)^{n}+\bigg(P-1-\sqrt{d}\bigg)^{n}\bigg) \\\\&=\frac{P-1}{\sqrt{d}}\sum_{j=0}^{n}\binom nj(P-1)^{n-j}\bigg((\sqrt{d})^j-(-\sqrt{d})^j\bigg) \\&\qquad\qquad+\sum_{j=0}^{n}\binom nj(P-1)^{n-j}\bigg((\sqrt{d})^j+(-\sqrt{d})^j\bigg) \\\\&=\sum_{k=1}^{(n+1)/2}\binom n{2k-1}(P-1)^{n-2k+2}2d^{k-1}+\sum_{k=0}^{(n-1)/2}\binom n{2k}(P-1)^{n-2k}2d^{k} \\\\&\equiv 2(P-1)d^{(n-1)/2}+2(P-1)^{n}\pmod n \\\\&\equiv 2(P-1)\bigg((P-1)^{n-1}-1\bigg)\pmod n \\\\&\equiv 0\pmod n\end{align}$$from which we get $$2^{n+1}S_{\delta(n)}\equiv 0\pmod n$$It follows from $\gcd(2^{n+1},n)=1$ that $$S_{\delta(n)}\equiv 0\pmod n$$
From the three cases above, the claim follows.$\quad\blacksquare$