Consider a finite set $S$, called the state space. Let $\Delta S$ be the set of all probability distributions on S. Consider a partition $\Pi$ of $S$, which is a collection of mutually disjoint subsets $E$ of $S$, whose union is $S$. Let $P \subseteq \Delta S$ be such that $p \in P$ implies $p(E)> 0$ for all $E \in \Pi$.
For each $E \in \Pi$ and $p \in P$, define $p_E = \frac{p}{p(E)}$ to be the Bayesian update of $p$ given $E$. Let $P_E$ be the set of all these conditionals of $P$ given $E$.
For each $r \in P$ and $p_E \in P_E$, define $p_E \otimes^E r \in \Delta S$ as follows: for each $F \subseteq S$, $p_E \otimes^E r (F) = r(E) p_E(F) + r(F \cap E^c)$. Hence, the choice of $p_E$ determines all probabilities given $E$, whereas $r$ determines all other probabilities.
Consider two properties:
a) Suppose convex set of probabilities P has the following property: For each $E \in \Pi$, for each $r \in P$ and each $p_E \in P_E$, we have $p_E \otimes^E r \in P$.
b) Suppose convex set P has the following property: it contains all probability measures that have the following property: take $r \in P$ and define $q = \underset{E \in \Pi}{\sum} r(E)p_E$, where $p_E \in P_E$. Then, $q \in P$. Note that $p_{E_1}$ and $p_{E_2}$ may not be conditionals of the same $p \in P$. Hence, $q$ is just the convex combination (with weights from $r$) of conditionals, one for each $E \in \Pi$.
The question is, if P satisfies property a), does it satisfy b)? And if it satisfies property b), does it satisfy property a)? Thanks!
Properties a) and b) are equivalent.
Your overloaded notation in b), in which the subscript on $p_E$ appears to denote both an index $E$ and the conditionalization on $E$, makes it difficult to express things clearly, so I'll write this as $p^E_E$, where $p^E$ is a distribution indexed by $E$ and $p^E_E$ is that distribution conditional on $E$, consistent with the notation you introduce at the beginning.
That b) implies a) is straightforward: In b), choose $p^E=r$ for all $E\in\Pi$ except $E=G$; then
$$ \sum_{E\in\Pi}r(E)p^E_E(F)=r(G)p^G_G(F)+\sum_{E\ne G}r(E)r_E(F)=r(G)p^G_G(F)+r(F\cap G^c)\;, $$
so any operation in a) can be mimicked by a single operation in b). In the other direction, you need several steps, one for each $E\in\Pi$. To mimick $\sum_{E\in\Pi}r(E)p^E_E$, start out with $r$ and in each step replace it by $p^E_E \otimes^E r$. Doing this for all $E\in\Pi$ transforms $r$ stepwise into $\sum_{E\in\Pi}r(E)p^E_E$.