For every prime $p$, there exists at least one index $k$ such that $$ \gcd(S_k,S_{k+1})=p. $$

\textbf{Theorem.} For every prime $p$, there exists at least one index $k$ such that
$$
\gcd(S_k,S_{k+1})=p.
$$

\textbf{Proof.}

Fix a prime $p$.

We use the Maynard--Tao theorem in the following form: for sufficiently large $K$, every admissible family of $K$ distinct fixed linear forms contains at least two primes for infinitely many values of the variable.

Choose such a $K$, and choose an admissible set
$$
\mathcal H=\{h_1<h_2<\cdots<h_K\}
$$
of nonnegative integers. Put
$$
H:=h_K.
$$

For example, if
$$
W:=\prod_{\substack{\ell\le K\\ \ell\ {\rm prime}}}\ell,
$$
we may take
$$
\mathcal H=\{0,W,2W,\ldots,(K-1)W\}.
$$

For every integer
$$
t\in\{0,1,\ldots,pH\}
\setminus\{ph_1,\ldots,ph_K\},
$$
choose three distinct primes
$$
\ell_{t,1},\ell_{t,2},\ell_{t,3}>pH,
$$
with no prime reused for two different values of $t$, and put
$$
m_t:=\ell_{t,1}\ell_{t,2}\ell_{t,3}.
$$

Thus the $m_t$ are pairwise coprime and
$$
\gcd(p,m_t)=1.
$$

For every such $t$, impose the congruence
$$
pa+t\equiv0\pmod{m_t}.
$$
Since the $m_t$ are pairwise coprime, the Chinese Remainder Theorem gives a simultaneous solution
$$
a\pmod M,
\qquad
M:=\prod_t m_t.
$$

Define
$$
L_i(n):=Mn+a+h_i,
\qquad 1\le i\le K.
$$

We claim that these linear forms are admissible.

First suppose that a prime $\ell$ divides $M$. Then $\ell$ divides a unique $m_t$. Since
$$
pa+t\equiv0\pmod\ell,
$$
we have
$$
p(a+h_i)\equiv ph_i-t\pmod\ell.
$$
Because $t\ne ph_i$ and
$$
|ph_i-t|\le pH<\ell,
$$
we obtain
$$
ph_i-t\not\equiv0\pmod\ell.
$$
Hence
$$
a+h_i\not\equiv0\pmod\ell
$$
for every $i$, and therefore
$$
L_i(n)\not\equiv0\pmod\ell
$$
for every $i$ and every $n$.

Now suppose that $\ell\nmid M$. Since $M$ is invertible modulo $\ell$, the residue classes of $n$ for which
$$
L_i(n)\equiv0\pmod\ell
$$
are an affine image of the residue classes occupied by
$$
h_1,\ldots,h_K.
$$
Since $\mathcal H$ is admissible, these do not cover all residue classes modulo $\ell$.

Thus
$$
\{L_1,\ldots,L_K\}
$$
is an admissible family of distinct linear forms.

By the Maynard--Tao theorem, there are infinitely many $n$ for which at least two of
$$
L_1(n),\ldots,L_K(n)
$$
are prime.

Fix one such sufficiently large $n$, and write
$$
N:=Mn+a.
$$
Choose two prime-valued forms
$$
q:=N+h_i,
\qquad
q':=N+h_j,
\qquad i<j,
$$
such that no $L_s(n)$ with
$$
i<s<j
$$
is prime.

Now consider
$$
pq
\qquad\text{and}\qquad
pq'.
$$
Both are semiprimes.

We claim that there is no semiprime strictly between them.

Let
$$
pq<x<pq'.
$$
Since $x$ is an integer, write
$$
x=pN+t.
$$
Then
$$
ph_i<t<ph_j.
$$

There are two cases.

If
$$
t\notin\{ph_1,\ldots,ph_K\},
$$
then $m_t$ was included in the CRT construction. Since $m_t\mid M$ and
$$
pa+t\equiv0\pmod{m_t},
$$
we have
$$
m_t\mid p(Mn+a)+t=x.
$$
But $m_t$ is a product of three distinct primes. Hence
$$
\Omega(x)\ge3,
$$
so $x$ is not a semiprime.

If instead
$$
t=ph_s
$$
for some $s$, then necessarily
$$
i<s<j.
$$
Therefore
$$
x=p(N+h_s)=pL_s(n).
$$
By the choice of $i$ and $j$, $L_s(n)$ is composite. Hence
$$
\Omega(L_s(n))\ge2,
$$
and therefore
$$
\Omega(x)\ge3.
$$
Again, $x$ is not a semiprime.

Thus there is no semiprime strictly between
$$
pq
\qquad\text{and}\qquad
pq'.
$$
Hence these are consecutive semiprimes. Therefore, for some index $k$,
$$
S_k=pq,
\qquad
S_{k+1}=pq'.
$$

Since $q$ and $q'$ are distinct primes,
$$
\gcd(S_k,S_{k+1})
=
\gcd(pq,pq')
=
p.
$$

Therefore, for the fixed prime $p$, there exists at least one index $k$ such that
$$
\boxed{\gcd(S_k,S_{k+1})=p.}
$$

Since $p$ was arbitrary, the result holds for every prime $p$.
$\boxed{}$