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
$$
\prod_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod_{v \in (\mathbf Z/(2^\beta))^\times} v\right)^{\varphi(n')} \equiv (\pm 1)^{\rm even} \equiv 1 \bmod 2^\beta.
$$
Best Answer
$\newcommand{floor}[1]{\left\lfloor #1 \right\rfloor}$
I doubt that this answer is still useful to you, since this question is one year old. Anyway, I'll leave the answer here, and maybe it will help someone (or not).
Your congruence is, modulo details, a special case of a congruence due to Voronoi. Let $k$ be an even positive integer, and write $B_k=\dfrac{N_k}{D_k}$ in lowest terms with $D_k>0$. Let $a$ and $n$ be coprime positive integers. Voronoi congruence states [2, Proposition 9.5.20]
that \begin{equation} \tag{1}\label{voronoin} (a^k-1)N_k\equiv D_kka^{k-1}\sum_{x=1}^{n-1} x^{k-1}\floor{\frac{ax}{n}} \quad(\mathrm{mod}\ n), \end{equation} where $\floor{x}$ is the usual floor function. Note that both sides are integers.
Let $n=p$ a prime number, and suppose $D_k$ is not divisible by $p$. Thus $p\ge5$ since $D_k$ is always divisible by $6$, by Clausen-von Staudt which states that [2, Corollary 9.5.15] $$D_k=\prod_{(\ell-1)\mid k}\ell,$$ the product being over all primes $\ell$ such that $\ell-1$ divides $k$. Then \eqref{voronoin} can be written as \begin{equation} \tag{2}\label{voronoip} (a^k-1)B_k\equiv ka^{k-1}\sum_{x=1}^{p-1} x^{k-1}\floor{\frac{ax}{p}} \quad(\mathrm{mod}\ p). \end{equation}
Let $c$ be a positive integer (coprime to $p$) such that $ac\equiv 1 \ (\mathrm{mod}\ p)$, and for simplicity write $a=c^{-1}$, being an inverse modulo $p$. (Obviously $c$ can be chosen such that $0<c<p$ as in your statement.) Multiplying both sides in \eqref{voronoip} by $c^k$ we obtain \begin{equation} \tag{3}\label{voronoic} (1-c^k)B_k\equiv k\sum_{x=1}^{p-1} x^{k-1}c\floor{\frac{c^{-1}x}{p}} \quad(\mathrm{mod}\ p). \end{equation} Now, $cc^{-1}=1+pN$ for some integer $N$, and we use this as follows to relate the summands with $h_c(x)$. For $1\le x\le p-1$, so that $x=\left(x \, \mathrm{mod}\, p\right)$, we have \begin{align*} c\floor{\frac{c^{-1}x}{p}} &=\frac{cc^{-1}x}{p}-\frac{c\left(c^{-1}x \; \mathrm{mod}\, p\right)}{p} =\frac{x}{p}+Nx-\frac{c\left(c^{-1}x \; \mathrm{mod}\, p\right)}{p}\\ &=h_c(x)+Nx-\frac{c-1}{2}, \end{align*} so that \eqref{voronoic} becomes \begin{equation} \tag{4}\label{voronoitemp} (1-c^k)B_k\equiv k\sum_{x=1}^{p-1} x^{k-1}h_c(x)+ kN\sum_{x=1}^{p-1}x^{k} -k\frac{c-1}{2}\sum_{x=1}^{p-1}x^{k-1} \quad(\mathrm{mod}\ p). \end{equation} To get rid of the last two sums, recall that for all $q$ not divisible by $p-1$ one has [1, Lemma 2.5.1] \begin{equation} \notag%\tag{4}\label{modp3} \sum_{x=1}^{p-1}x^{q}\equiv 0 \quad(\mathrm{mod}\ p). \end{equation} Now we use your hypothesis $2\le k\le p-3$; then both $k-1$ and $k$ are not divisible by $p-1$, and the above implies that the last two sums in \eqref{voronoitemp} are zero. Also under this hypothesis, Clausen-von Staudt implies that $p$ does not divide $D_k$, which was used in \eqref{voronoip}. If moreover $c^k \not\equiv 1 \pmod p$ we finally arrive to your congruence \begin{equation} \notag%\tag{4}\label{modp3} B_k\equiv \frac{k}{1-c^k}\sum_{x=1}^{p-1} x^{k-1}h_c(x)\quad(\mathrm{mod}\ p). \end{equation}
[1] Cohen, Henri Number theory. Vol. I. Tools and Diophantine equations. Graduate Texts in Mathematics, 239. Springer, New York, 2007.
[2] Cohen, Henri Number theory. Vol. II. Analytic and modern tools. Graduate Texts in Mathematics, 240. Springer, New York, 2007.
(The original article by Voronoi can be accesed here. His congruence appears as Corollary IV in page 146.)