Twin primes satisfy the congruence?

By Wilson's theorem, $(p-1)!\equiv-1\bmod p$ if and only if $p$ is a prime. So assume that $6n-1$ and $6n+1$ are primes. Then $(6n-2)!\equiv-1\bmod(6n-1)$, and $(6n)!\equiv-1\bmod(6n+1)$. The second congruence can be rewritten as $(6n-2)!\equiv-1(-1)^{-1}(-2)^{-1}=-2^{-1}\bmod(6n+1)$.

Multiplying both congruences by $4$ yields

$$4(6n-2)!\equiv-4\bmod(6n-1)\;,\\4(6n-2)!\equiv-2\bmod(6n+1)\;.$$

Now we just have to check that these are indeed the residues of $-3(1+2n)$.

Writing your twin primes as $(p, p+2)$, your observation can be rewritten as:

$4(p-1)! = -4-p$ (mod $p^2 + 2p$).

This is known as Clement's Theorem.

Note, by the way, that this is an if and only if statement: the above relation is satisfied if and only if $(p, p+2)$ are twin primes.

(And yes, Wilson's Theorem is typically used to prove Clement's Theorem. Cf. Theorem 2 here.)