I will show the two results are non-superficially related by showing one of them implies the other: the classification of moduli $n \geq 2$ for which the unit group $(\mathbf Z/(n))^\times$ is cyclic implies Gauss' generalization of Wilson's theorem.

The proof is presented in three steps. All the basic ideas are present in the case of odd $n$, which doesn't need the third step. Handling even $n$ is mostly a matter of tedious details.

Step 1: For $n \geq 2$, if $(\mathbf Z/(n))^\times$ is cyclic, then $\prod_{u \in (\mathbf Z/(n))^\times} u \equiv -1 \bmod n$.

Proof: The result is obvious for $n = 2$, so we can take $n \geq 3$, which implies $\varphi(n)$ is even. Let $g$ be generator of $(\mathbf Z/(n))^\times$. Then
$$
\prod_{u \in (\mathbf Z/(n))^\times} u =
\prod_{0 \leq k \leq \varphi(n)-1} g^k =
g^{\varphi(n)(\varphi(n)-1)/2} \bmod n.
$$
Since $g$ has order $\varphi(n)$, which is even,
$g^{\varphi(n)/2}$ has order 2 in $(\mathbf Z/(n))^\times$, so it must be $-1$ (the only element of order $2$ in the cyclic group $(\mathbf Z/(n)^\times$). Thus
$$
g^{\varphi(n)(\varphi(n)-1)/2} =
\left(g^{\varphi(n)/2}\right)^{\varphi(n)-1} =
(-1)^{\varphi(n)-1} = -1 \bmod n
$$
since $\varphi(n)-1$ is odd.

Step 1 covers the cases $n = 2$, $4$, $p^\alpha$, and $2p^\alpha$ where $p$ is an odd prime and $\alpha \geq 1$. The next two steps handle the remaining $n$.

Step 2: For odd $n > 1$ that is not a prime power, $\prod_{u \in (\mathbf Z/(n))^\times} u \equiv 1 \bmod n$.

Proof: To prove that product over units in $(\mathbf Z/(n))^\times$ is $1$, it suffices to show for each prime power $p^\alpha\mid\mid n$ that the product is
$1 \bmod p^\alpha$ (then use the Chinese remainder theorem).

Write $n = p^\alpha m$, so $\gcd(p^\alpha,m) = 1$.
The natural reduction homomorphism $(\mathbf Z/(n))^\times \to (\mathbf Z/(p^\alpha))^\times$ is surjective, so each unit mod $p^\alpha$ is the reduction of $\varphi(n)/\varphi(p^\alpha)$ units mod $n$, and
$\varphi(n)/\varphi(p^\alpha) = \varphi(m)$. Thus
$$
\prod_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod_{v \in (\mathbf Z/(p^\alpha))^\times} v\right)^{\varphi(m)} \bmod p^\alpha.
$$
The group $(\mathbf Z/(p^\alpha))^\times$ is cyclic, so by Step 1 the product over $v$ on the right side is $-1 \bmod p^\alpha$ and the exponent $\varphi(m)$ is even because $m \geq 3$ (this is where we use the fact that $n$ is odd and not a prime power), so the right side of the displayed congruence above is $1 \bmod p^\alpha$.

Step 3: For even $n > 1$ that is not $2$, $4$, or $2p^\alpha$ for an odd prime $p$, $\prod_{u \in (\mathbf Z/(n))^\times} u \equiv 1 \bmod n$.

Write $n = 2^\beta n'$ for $\beta \geq 1$ and odd $n' \geq 1$.
We describe these $n$ in three ways: (i) $n = 2^\beta$ for $\beta \geq 3$, (ii) $n = 2n'$ where $n' > 1$ is not a prime power, or (iii) $n = 2^\beta n'$ where $\beta \geq 2$ and $n' \geq 3$.

(i): $n = 2^\beta$ for $\beta \geq 3$. Show by induction on $\beta$ that the solutions of $x^2 \equiv 1 \bmod 2^\beta$ are $x \equiv \pm 1, \pm(1+ 2^{\beta-1}) \bmod 2^\beta$, which are all distinct since $\beta \geq 3$. Therefore
$$
\prod_{u \in (\mathbf Z/(2^\beta))^\times} u \equiv
(-1)(1+2^{\beta-1})(-(1+2^{\beta-1})) \equiv 1 \bmod 2^\beta.
$$

(ii): $n = 2n'$ where $n' > 1$ is not a prime power. We will argue as in Step 2.
The natural reduction homomorphism $(\mathbf Z/(n))^\times \to (\mathbf Z/(n'))^\times$ is surjective, so each unit mod $n'$ is the reduction of $\varphi(n)/\varphi(n')$ units mod $n$. Since
$\varphi(n) = \varphi(2n') = \varphi(2)\varphi(n') = \varphi(n')$,
$\varphi(n)/\varphi(n') = 1$, so
$$
\prod_{u \in (\mathbf Z/(n))^\times} u \equiv
\prod_{v \in (\mathbf Z/(n'))^\times} v\bmod n'.
$$
Since $n' > 1$ is odd and not a prime power, the product on the right side of the displayed congruence is $1 \bmod n'$ by Step 2.
So the product on the left side of the displayed congruence is $1 \bmod n'$. It is also $1 \bmod 2$ since units mod $n$ are odd.
Therefore $\prod_{u \in (\mathbf Z/(n))^\times} u$ is $1 \bmod n'$ and $1 \bmod 2$, which makes it $1 \bmod n$.

(iii) $n = 2^\beta n'$ where $\beta \geq 2$ and $n' \geq 3$.
Using the same method as in Step 2, to show $\prod_{u \in (\mathbf Z/(n))^\times} u$ is $1 \bmod n$, it suffices to show the product is $1 \bmod 2^\beta$ and $1 \bmod n'$.

First we show the product is $1 \bmod n'$.
The natural reduction homomorphism $(\mathbf Z/(n))^\times \to (\mathbf Z/(n'))^\times$ is surjective, so each unit mod $n'$ is the reduction of $\varphi(n)/\varphi(n')$ units mod $n$, and
$\varphi(n)/\varphi(n') = \varphi(2^\beta)$. Thus
$$
\prod_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod_{v \in (\mathbf Z/(n'))^\times} v\right)^{\varphi(2^\beta)} \bmod n'.
$$
On the right side, the product over units modulo $n'$
is $-1 \bmod n'$ if $n'$ is a prime power (Step 1) and
it is $1 \bmod n'$ if $n'$ is not a prime power (Step 2). Since $\varphi(2^\beta)$ is even,
$$
\left(\prod_{v \in (\mathbf Z/(n'))^\times} v\right)^{\varphi(2^\beta)} \equiv (\pm 1)^{\rm even} \equiv 1 \bmod n'.
$$

To show the product is $1 \bmod 2^\beta$, swap the roles of $2^\beta$ and $n'$ in the previous argument to get
$$
\prod_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod_{v \in (\mathbf Z/(2^\beta))^\times} v\right)^{\varphi(n')} \equiv (\pm 1)^{\rm even} \equiv 1 \bmod 2^\beta.
$$