I am looking for occurrences in literature of the following property of a formal language $\mathcal L$ over an alphabet $\Sigma$
For any quadruple of words $a,b,c,d\in\Sigma^*$, if $ac,bc,ad\in\mathcal L$, then $bd\in\mathcal L$.
The motivation of this property comes from a non-commutative version of quantum logic. Intuitively, one can interpret the words of such a language as "all possible ways of achieving the same result" in this context.
We discussed your question with @M.Monet in the case of regular languages. It turns out that, unless we're mistaken, your condition is precisely equivalent to saying that the language is recognized by a bideterministic automaton. An automaton is bideterministic if it is deterministic (so in particular has a unique initial state), if it has a unique final state, and if the reverse of the automaton is also deterministic. (The notion exists under different names, e.g., injective automata, zero-reversible acceptors...) I will call a language bideterministic if it is accepted by such an automaton. It is known that not all languages are bideterministic, but that if a language is bideterministic then the minimal deterministic automaton for the language will be bideterministic. See for instance Pin, On reversible automata, https://hal.science/hal-00019977.
What follows is a proof of the fact that a regular language satisfies your condition iff it is bideterministic. I don't know whether this characterization has been stated before somewhere in the literature on bideterministic automata.
Let me first show that a language $L$ accepted by a bideterministic automaton $A$ satisfies your condition. Assume that $ac \in L$, let $q_0$ be the initial state of $A$, $q_f$ the final state, and $q$ the state reached between $a$ and $c$. Further assume that $ad \in L$. Hence there is a path $\pi'$ labeled $d$ from $q$ to $q_f$. Now assume $bc \in L$. We know that the run of $bc$ in the automaton must be reaching state $q$ between $b$ and $c$, as otherwise we have two reverse paths when reading the reverse of $c$ backwards from $q_f$ that lead to two different states, and this would violate the reverse determinism requirement. Thus, there is a path $\pi$ labeled $b$ from $q_0$ to $q$. We conclude from $\pi$ and $\pi'$ that the word $bd$ is also accepted by the automaton so your requirement is satisfied.
To show a converse, let $L$ be a regular language. Let us consider its minimal deterministic automaton, and assume that we have trimmed the minimal automaton: call $A$ the result. Let me first show that if $A$ has two final states then $L$ does not satisfy your condition. We prove this by contradiction. Let $q_f$ and $q_f'$ be two different final states. As $A$ is minimal, the language recognized from $q_f$ and $q_f'$ must be different (Myhill–Nerode): up to exchanging them, we assume without loss of generality that there is a word $d$ such that $d$ is accepted from $q_f$ and $d$ is not accepted from $q_f'$. Now, as $A$ is minimal, $q_f$ and $q_f'$ are accessible. Hence, let $a$ be a word going from the initial state to $q_f$, and let $b$ be a word going from the initial state to $q_f'$. Now, take $c = \epsilon$. We know that $ac = a$ and $bc = b$ are both accepted, and $ad$ is accepted (because reading $a$ from the initial state leads to $q_f$ from which we accept $d$). Hence your condition would imply that $bd$ is accepted. But this is a contradiction: reading $b$ from the initial state leads to $q_f'$ from which we reject $d$ by assumption.
Hence, $A$ must have a single final state $q_f$. Let me show that the reverse of $A$ must be deterministic. The proof is by contradiction and is similar to the previous paragraph. Assume that there is a violation of reverse determinism, i.e., a state $q$ and letter $x$ for which there are two distinct predecessors $q_1$ and $q_2$. As $A$ is minimal, we know that the language accepted from $q_1$ and $q_2$ cannot be the same. So, up to exchanging $q_1$ and $q_2$, without loss of generality assume that there is a word $d$ which is accepted from $q_1$ and rejected from $q_2$. Now, $A$ is trimmed so $q$ is co-accessible. Hence, let $c'$ be a word accepted from $q$: thus the word $c = xc'$ is accepted from both $q_1$ and $q_2$. As $A$ is trimmed, $q_1$ and $q_2$ are accessible, so let $a$ be a word going from the initial state to $q_1$, and $b$ be a word going from the initial state to $q_2$. We can now see that $ac$ is accepted (via $q_1$) and $bc$ is accepted (via $q_2$) and $ad$ is accepted (via $q_1$) but $bd$ is rejected (the run goes via $q_2$ from which we reject $d$. So we have a contradiction. This together with the previous paragraph establishes the reverse direction of the equivalence.
