Suppose that there is some underlying binary state in $\Omega = \{ L,R \}$. You observe a signal $s$ from some finite set $S$, with $\vert S \vert \ge 2$. The probability of observing each signal depends on the state. Denote $\pi_\omega(s) = \Pr(s\vert \omega)$, $\omega \in \Omega$. Call the collection $\{ \pi_\omega (s) \}_{\omega\in\Omega,s\in S}$ a signal distribution.
We can calculate the likelihood ratio of the signal $s$ via
$$ \rho(s) = \frac{\pi_L(s)}{\pi_R (s)}. $$
Given a collection of likelihood ratios $\{ \rho(s) \}_{s \in S}$, when does there exist a signal distribution whose likelihood ratios are exactly the ones specified?
Formally, this question is equivalent to asking when there is a solution to the following system of equalities:
\begin{align*} \sum_{s\in S} \pi_\omega(s) &= 1, \; \forall\omega \in \Omega \\ \rho(s) \pi_R(s) &= \pi_L(s), \; \forall s \in S \\ \end{align*}
such that $0 \le \pi_\omega (s) \le 1$ for all $\omega$ and $s$.
If we ignore the requirement that the probabilities must be contained in $[0,1]$, then the resulting linear system has a large number of degrees of freedom, and it becomes easy to guarantee the existence of a solution. Indeed, there will be many solutions when $\vert S \vert > 2$. However, it isn't clear to me how to check if any of the solutions are all in $[0,1]$.
One obvious necessary condition for the solution to be a pair of probability distributions is that not all of the $\rho(s)$'s can be strictly greater than $1$, nor can all of them be strictly less than $1$, otherwise the system of equalities would be inconsistent. Simple calculations show that this necessary condition is sufficient for the case when $\vert S \vert = 2$. These calculations become much more involved for $\vert S \vert > 2$. Can anyone suggest a simpler approach?
The necessary condition you found is indeed also sufficient for $|S|\gt2$ (except I believe you misstated it slightly – unless all ratios are $1$, there must be at least one ratio on each side of $1$ (whereas your formulation allows cases where some ratios are $1$ and the other are all on one side of $1$)).
To see this, find two ratios on either side of $1$ and set the probabilities for all other elements of $S$ to $0$. If you don't want to count it as satifying the ratio constraints if both probabilities are zero, you can assign arbitrarily small probabilities instead. Then you just have to modify your proof for the case $|S|=2$ such that it also covers the case where the probabilities add up to less than $1$.