I think it’s true that if $R$ is any commutative, unital ring, then if the formal Laurent series ring $R((t))$ is isomorphic to a Cartesian product of two rings $R((t)) \cong S_1 \times S_2$, then both factors are themselves Laurent series rings. That is, there must be rings $R_1$ and $R_2$ such that $S_i \cong R_i((t))$.
I can easily show this for the formal power series ring $R[[t]]$ using idempotents: equivalent to the existence of $S_1$ and $S_2$ are a complementary pair of idempotents $e+f=1$. If they are both contained in $R[[t]]$, then their degree-$0$ coefficients $e_0$ and $f_0$ are also idempotent. In particular $e_0 + f_0 = 1$, so $R[[t]] = (e_0R)[[t]] \times (f_0R)[[t]]$.
However, the existence of negative-degree coefficients is really complicating my attempts to prove the claim for $R((t))$ in the same way.
Does anyone have any ideas for how to continue, or know any alternative proofs, or a counterexample? Geometrically it seems like it should be true—I don’t see why $\mathrm{Spec}(R((t)))$ should be allowed to have more disjoint components than $\mathrm{Spec}(R)$, but I’m really stuck here.