Excluding 1 being neither prime nor composite, I found the following closed forms for infinite products of composite numbers (happy to share how I derived these).
$$\displaystyle \prod _{composites}^{\infty } \left( \dfrac{{c}^{2}} {{c}^{2}-1}\right) = \frac{12}{\pi^2}$$
and
$$\displaystyle \prod _{composites}^{\infty } \left( \dfrac{{c}^{2}} {{c}^{2}+1}\right) = \frac{30}{\pi \sinh(\pi)}$$
Easy to see that multiplying both gives:
$$\displaystyle \prod _{composites}^{\infty } \left( \dfrac{{c}^{4}} {{c}^{4}-1}\right) = \frac{360}{\pi^3 \sinh(\pi)}$$
The latter is a product that very rapidly converges.
Are there any other closed forms for these "Euler-type" products of composites known?
Still not sure that I understand the question clearly, but for any integer values $k>1$,
Products of the form: $$\prod_{c\text{ composite}}\frac{c^k}{c^k-1}$$
Can be evaluated in terms of a product of Gamma functions evaluated at roots of unity, a zeta constant, and a suitable rational scalar.
For example the case for when $k=2n$ is an even integer is given by,
$$\prod_{c\text{ composite}}\frac{c^{2n}}{c^{2n}-1}=\frac{(2n)!(4n)(-\pi i)^{n-1}}{\text{B}_{2n}(2\pi)^{2n}}\prod_{m=1}^{n-1}\csc(\pi e^{i\pi m/n})$$
For all natural numbers $n$, where $\text{B}_{n}$ are the Bernoulli numbers.
And for the special case $n=1$ the product on the right is simply assumed to be empty.
The key idea in obtaining this and similar products is to take note upon two ideas,
One, that given:
$$\prod_{n=2}^\infty\frac{n^s}{n^s-1}=f(s)$$
Can be evaluated in terms of the gamma function when $\text{s}$ is an integer greater then one.
And two, that through the use of the Euler product representation of the Riemann zeta function:
$$\prod_{p\text{ prime}}\frac{p^s}{p^s-1}=\zeta(s)$$
We can take the corresponding ratios of the two functions so that the primes occurring in $f$ are canceled out by the primes occurring in $\zeta$, leaving us with a product ranging over the composite numbers, that is:
$$\prod_{c\text{ composite}}\frac{c^s}{c^s-1}=\frac{f(s)}{\zeta(s)}$$