If I am given the data of the marginals $P(A,B)$ and $P(B,C)$ together with the promise that $A$ and $C$ are independent, i.e. $P(A,C)=P(A)P(C)$, then, is there a way to deduce the full distribution $P(A,B,C)$ from this data?
Or contrary, do there exist multiple distributions $P$ and $Q$ with the same marginals on $A,B$ and $B,C$ and for which $A$ and $C$ are independent but with $P\neq Q$?
Edit
Thanks to the answer below! I also realized another simple example, namely $P$ over $A,B,C=0,1$ \begin{align} P(A,B,C)= \begin{cases} \frac{1}{4},A+B+C\text{ even}\newline 0,\text{else} \end{cases} \end{align} I think this is the same as your example just over the field $\mathbb{F}_2$ instead of the reals. Then all two variable reduced marginals equal the uniform distribution. This implies (a) that every pair of variables is mutually independent and (b) based on this data $P$ cannot be distinguished from the uniform distribution.
Luckily I realized that in the classes I consider these distributions can not be realized. More precisely, I realized that I do not only have independence of $A$ and $B$, but also conditional independence: \begin{align} P(A,C\mid B)=P(A\mid B)P(C\mid B), \end{align} such that I can obtain all information via
\begin{align} P(A,B,C)=P(B)P(A,C\mid B)=P(B)P(A\mid B)P(C\mid B)=\frac{P(A,B)P(B,C)}{P(B)} \end{align}
If it can be shown that the information is not deducible when B also happens to be an independent event (so that all 3 events are independent of each other), that should be conclusive.
Let $a = p(A), b = p(B), c = P(C).$
In effect, you are given $(ab)$ and $(bc)$ and want to deduce from this $(abc)$.
You can deduce $(ab^2c)$ but not $(abc)$.
Take 2 examples:
$a = c = (1/2), b = (1/\sqrt{2}) \implies ab = bc = (1/2\sqrt{2}), abc = (1/4\sqrt{2}).$
$a = c = (1/2\sqrt{2}), b = 1 \implies ab = bc = (1/2\sqrt{2}), abc = (1/8).$