On Shanks' quartic approximation $\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)$

Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then

$$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$

As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressions above are genuinely units. I wrote them in this form to conform with the previous example of the OP) The first unit (involving $a$) generates a degree four extension whose Galois closure has Galois group $D_8$, but the latter generates a Galois extension with Galois group $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$.

For those playing at home, this allows for the extra simplification

$$b + \sqrt{b^2 + 1} = (1 + \sqrt{2})^2 \cdot \frac{7 + \sqrt{53}}{2} .$$

The following is not a direct answer to your question, but is rather too long for a comment.

This is basically Ramanujan's approximation $$\pi \approx \frac{24}{\sqrt{n}}\log(2^{1/4}g_{n}) = \frac{6}{\sqrt{n}}\log (2g_{n}^{4}) = \frac{6}{\sqrt{n}}\log (2u)\tag{1}$$ where $g_{n}$ is Ramanujan's class invariant given by $$g_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\tag{2}$$ Here $n = 3502 = 2 \times 17 \times 103$ and it is known that for many values of $n$ the class invariant $g_{n}$ is a unit so that that $u = g_{n}^{4}$ is also a unit. The value of $u$ given in the question is clearly a unit.

Also raising both sides of $(2)$ to 24th power and doing some simpification we get $$(2u)^{6} + 24 = 64g_{n}^{24} + 24 = e^{\pi\sqrt{n}} + 276e^{-\pi\sqrt{n}} + \cdots = e^{\pi\sqrt{n}}(1 + 276e^{-2\pi\sqrt{n}} + \cdots)\tag{3}$$ and hence on taking logs we get $$\log\{(2u)^{6} + 24\} = \pi\sqrt{n} + \log(1 + 276e^{-2\pi\sqrt{n}}) \approx \pi\sqrt{n} + 276e^{-2\pi\sqrt{n}}$$ and thus $$\pi \approx \frac{1}{\sqrt{n}}\log\{(2u)^{6} + 24\} - \frac{276}{\sqrt{n}}e^{-2\pi\sqrt{n}}\tag{4}$$ The error is of the order of $e^{-2\pi\sqrt{n}}$ and in formula $(1)$ it is of the order $e^{-\pi\sqrt{n}}$ (as evident from infinite product $(2)$). Putting $n = 3502$ we see that error in formula $(1)$ is of the order $10^{-81}$ and that in formula $(4)$ is of the order $10^{-161}$. See Ramanujan's paper "Modular Equations and Approximations to $\pi$".