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 two claims are true.
Proof for the first claim :
It can be proven by induction on $i$ that
$$S_i=2^{-a^{i+1}}(s^{a^{i+1}}+t^{a^{i+1}})\tag1$$
where $s:=c-\sqrt{c^2-4},t:=c+\sqrt{c^2-4}$.$\ \ $ (The proof can be seen in this answer.)
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 $(2)(3)(4)(5)$ that
$$\begin{align}2^{(2a-2)(N-1)}\cdot S_{p-1}&=2^{-2a+1}\bigg(s((\sqrt{c+2}-\sqrt{c-2})^{N})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t((\sqrt{c+2}+\sqrt{c-2})^{N})^{2(a-1)}\bigg)
\\\\&=2^{-2a+1}\bigg(s(C\sqrt{c+2}- D\sqrt{c-2})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t(C\sqrt{c+2}+ D\sqrt{c-2})^{2(a-1)}\bigg)
\\\\&\equiv 2^{-2a+1}\bigg(s(\sqrt{c+2}+\sqrt{c-2})^{2(a-1)}
\\&\qquad\qquad\qquad\qquad+t(\sqrt{c+2}-\sqrt{c-2})^{2(a-1)}\bigg)\pmod N
\\\\&= 2^{-2a+1}(s\cdot 2^{a-1}t^{a-1}+t\cdot 2^{a-1}s^{a-1})
\\\\&= 2^{-(a-2)}(s^{a-2}+t^{a-2})
\\\\&=P_{a-2}(c)\end{align}$$
It follows from $2^{N-1}\equiv 1\pmod N$ that
$$S_{p-1}\equiv P_{a-2}(c)\pmod{M_p(a)}\quad\blacksquare$$
Proof for the second claim :
Let $N:=N_p(b)= \frac{b^p+1}{b+1}$. Then, from $(1)$, we get, similarly as above,
$$\begin{align}2^{2(b+1)(N-1)}\cdot S_{p-1}&=2^{(b+1)N-2b-1}(s^{(b+1)N-1}+t^{(b+1)N-1})
\\\\&=2^{-2b-3}\bigg(t\cdot ((\sqrt{c+2}-\sqrt{c-2})^N)^{2(b+1)}
\\&\qquad\qquad\qquad\qquad +s\cdot ((\sqrt{c+2}+\sqrt{c-2})^N)^{2(b+1)}\bigg)
\\\\&=2^{-2b-3}\bigg(t\cdot (C\sqrt{c+2}-D\sqrt{c-2})^{2(b+1)}
\\&\qquad\qquad\qquad\qquad +s\cdot (C\sqrt{c+2}+D\sqrt{c-2})^{2(b+1)}\bigg)
\\\\&\equiv 2^{-2b-3}\bigg(t\cdot (\sqrt{c+2}+\sqrt{c-2})^{2(b+1)}
\\&\qquad\qquad\qquad\qquad +s\cdot (\sqrt{c+2}-\sqrt{c-2})^{2(b+1)}\bigg)\pmod N
\\\\&=2^{-2b-3}(t\cdot 2^{b+1}t^{b+1}+s\cdot 2^{b+1}s^{b+1})
\\\\&=2^{-(b+2)}(s^{b+2}+t^{b+2})
\\\\&=P_{b+2}(c)
\end{align}$$
from which
$$S_{p-1}\equiv P_{b+2}(c)\pmod{N_p(b)}$$
follows.$\quad\blacksquare$
Best Answer
The two claims are true.
Proof for Claim 1 :
Let us prove by induction that $$S_i\equiv 2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})\pmod M\tag1$$ where $s:=a-\sqrt{a^2-4},t=a+\sqrt{a^2-4}$.
$(1)$ holds for $i=0$ since $$S_0\equiv P_{kb}(a)=2^{-kb}(s^{kb}+t^{kb})\pmod M$$
Supposing that $(1)$ holds for $i$ gives $$\begin{align}S_{i+1}&=P_b(S_i) \\\\&\equiv P_b(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}}))\pmod M \\\\&=2^{-b}\cdot\bigg(\bigg(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})-\sqrt{(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}}))^2-4}\bigg)^b \\&\qquad\qquad+\bigg(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})+\sqrt{(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}}))^2-4}\bigg)^b\bigg) \\\\&=2^{-b}\cdot\bigg(\bigg(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})-\sqrt{(2^{-kb^{i+1}}(t^{kb^{i+1}}-s^{kb^{i+1}}))^2}\bigg)^b \\&\qquad\qquad +\bigg(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})+\sqrt{(2^{-kb^{i+1}}(t^{kb^{i+1}}-s^{kb^{i+1}}))^2}\bigg)^b\bigg) \\\\&=2^{-b}\cdot\bigg(\bigg(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})-2^{-kb^{i+1}}(t^{kb^{i+1}}-s^{kb^{i+1}})\bigg)^b \\&\qquad\qquad+\bigg(2^{-kb^{i+1}}(s^{kb^{i+1}}+t^{kb^{i+1}})+2^{-kb^{i+1}}(t^{kb^{i+1}}-s^{kb^{i+1}})\bigg)^b\bigg) \\\\&=2^{-kb^{i+2}}(s^{kb^{i+2}}+t^{kb^{i+2}})\qquad\square\end{align}$$
From $(1)$, we get, using $2(a\pm\sqrt{a^2-4})=(\sqrt{a+2}\pm\sqrt{a-2})^2$, $$\begin{align}S_{n-1}&\equiv 2^{-(M+c)}(s^{M+c}+t^{M+c})\pmod M \\\\&=2^{-2M-2c}\bigg((\sqrt{a+2}-\sqrt{a-2})^{2M+2c} \\&\qquad\qquad\qquad +(\sqrt{a+2}+\sqrt{a-2})^{2M+2c}\bigg)\tag2\end{align}$$
Here, we have, by the binomial theorem, $$\begin{align}&(\sqrt{a+2}\pm\sqrt{a-2})^{M} \\\\&=\sum_{k=0}^{M}\binom Mk(\sqrt{a+2})^{M-k}(\pm\sqrt{a-2})^k \\\\&=\sqrt{a+2}\sum_{j=0}^{(M-1)/2}\binom{M}{2j}(a+2)^{(M-2j-1)/2}(a-2)^{j} \\&\qquad\qquad \pm\sqrt{a-2}\sum_{j=1}^{(M+1)/2}\binom{M}{2j-1}(a+2)^{(M-2j+1)/2}(a-2)^{j-1}\end{align}$$
So, there are integers $C,D$ such that $$(\sqrt{a+2}\pm\sqrt{a-2})^{M}=C\sqrt{a+2}\pm D\sqrt{a-2}$$ where $$C=\sum_{j=0}^{(M-1)/2}\binom{M}{2j}(a+2)^{(M-2j-1)/2}(a-2)^{j}\equiv \left(\frac{a+2}{M}\right)\equiv 1\pmod M$$ and $$D=\sum_{j=1}^{(M+1)/2}\binom{M}{2j-1}(a+2)^{(M-2j+1)/2}(a-2)^{j-1}\equiv \left(\frac{a-2}{M}\right)\equiv -1\pmod M$$
From $(2)$, we get $$\begin{align}2^{2(M-1)}\cdot S_{n-1}&\equiv 2^{-2c-2}\left((\sqrt{a+2}-\sqrt{a-2})^{2M+2c}+(\sqrt{a+2}+\sqrt{a-2})^{2M+2c}\right)\pmod M \\\\&=2^{-2c-2}\bigg((\sqrt{a+2}-\sqrt{a-2})^{2c}(C\sqrt{a+2}- D\sqrt{a-2})^2 \\&\qquad +(\sqrt{a+2}+\sqrt{a-2})^{2c}(C\sqrt{a+2}+ D\sqrt{a-2})^2\bigg) \\\\&\equiv 2^{-2c-2}\bigg((\sqrt{a+2}-\sqrt{a-2})^{2c}(\sqrt{a+2}+\sqrt{a-2})^2 \\&\qquad +(\sqrt{a+2}+\sqrt{a-2})^{2c}(\sqrt{a+2}-\sqrt{a-2})^2\bigg)\pmod M \\\\&=2^{-2c+2}\bigg((\sqrt{a+2}-\sqrt{a-2})^{2c-2}+(\sqrt{a+2}+\sqrt{a-2})^{2c-2}\bigg) \\\\&=2^{-(c-1)}(s^{c-1}+t^{c-1}) \\\\&=P_{c-1}(a)\end{align}$$ It follows from $2^{2(M-1)}\equiv 1\pmod M$ that $$S_{n-1}\equiv P_{c-1}(a)\pmod M$$
Proof for Claim 2 :
From $(1)$, we get, similarly as above, $$\begin{align}2^{2(N-1)}\cdot S_{n-1}&\equiv 2^{N+c-2}(s^{N-c}+t^{N-c})\pmod N \\\\&=2^{2c-2}((\sqrt{s+2}-\sqrt{s-2})^{2N-2c}+(\sqrt{s+2}+\sqrt{s-2})^{2N-2c}) \\\\&=2^{2c-2}\bigg((\sqrt{s+2}-\sqrt{s-2})^{2N}\bigg(\frac{\sqrt{s+2}+\sqrt{s-2}}{4}\bigg)^{2c} \\&\qquad\qquad +(\sqrt{s+2}+\sqrt{s-2})^{2N}\bigg(\frac{\sqrt{s+2}-\sqrt{s-2}}{4}\bigg)^{2c}\bigg) \\\\&\equiv 2^{-2c-2}((\sqrt{s+2}-\sqrt{s-2})^2(\sqrt{s+2}+\sqrt{s-2})^{2c} \\&\qquad\qquad +(\sqrt{s+2}+\sqrt{s-2})^2(\sqrt{s+2}-\sqrt{s-2})^{2c})\pmod N \\\\&=2^{-2c+2}((\sqrt{s+2}+\sqrt{s-2})^{2c-2}+(\sqrt{s+2}-\sqrt{s-2})^{2c-2}) \\\\&=2^{-(c-1)}(s^{c-1}+t^{c-1}) \\\\&=P_{c-1}(a)\end{align}$$ It follows from $2^{2(N-1)}\equiv 1\pmod N$ that $$S_{n-1}\equiv P_{c-1}(a)\pmod N$$