Here is a framework that might be applicable.
There are two related entities. There are $T_i$ observations for entity: $Y_i = (y_{i1}, \ldots, y_{iT_i})$ for $i \in \{1,2\}$. Each of the two entities depends on an unknown parameter $\mu_i$: $p(Y_i|\mu_i)$. If the observations for entity $i$ were iid (independent and identically distributed) then
\begin{equation}
p(Y_i|\mu_i) = \prod_{t=1}^{T_i} p(y_{it}|\mu_i) .
\end{equation}
Given this setup, the only possible connection between $\mu_2$ and $Y_1$ (for example) is through the joint prior for $(\mu_1,\mu_2)$. In other words, we can learn about $\mu_2$ from $Y_1$ if we believe that we can learn about $\mu_2$ from $\mu_1$. We can complete the connection in two steps. First,
\begin{equation}
p(\mu_2|Y_1,\mu_1) = \frac{p(Y_2|\mu_2)\,p(\mu_2|\mu_1)}{p(Y_2|\mu_1)} .
\end{equation}
Second, given $p(\mu_1|Y_1)$ we have
\begin{equation}
p(\mu_2|Y_1,Y_2) = \int p(\mu_1|Y_2,\mu_1)\,p(\mu_1|Y_1)\,d\mu_1 .
\end{equation}
We can arrive at the same place via a different route. First compute
\begin{equation}
p(\mu_2|Y_1) = \int p(\mu_2|\mu_1)\,p(\mu_1|Y_1)\,d\mu_1
\end{equation}
and second
\begin{equation}
p(\mu_2|Y_1,Y_2) = \frac{p(Y_2|\mu_2)\,p(\mu_2|Y_1)}{p(Y_2|Y_1)} .
\end{equation}
(Note that the sample information $Y_2$ is contained in the likelihood and the non-sample information $Y_1$ is contained in the prior.)
Either way, you would need to come up with $p(\mu_2|\mu_1)$, the distribution for $\mu_2$ if you knew $\mu_1$, and $p(\mu_1|Y_1)$, which is what you know about $\mu_1$ from its own dataset.
Updating
Suppose we get an additional observation for entity 2 so that $Y_2' = (Y_2, y_2')$. Then
\begin{equation}
p(\mu_2|Y_1,Y_2') = \frac{p(y_2'|\mu_2)\,p(\mu_2|Y_1,Y_2)}{p(y_2|Y_1,Y_2)} .
\end{equation}
Nuisance parameters.
If there are nuisance parameters, they can be integrated out. For example,
\begin{equation}
p(Y_i|\mu_i) = \int p(Y_i|\mu_i,\sigma_i^2)\,p(\sigma_i^2|\mu_i)\,d\sigma_i^2 .
\end{equation}
To flesh this example out a bit, suppose
\begin{equation}
p(Y_i|\mu_i,\sigma_i) = \prod_{t=1}^{T_i} \textsf{N}(y_{it}|\mu_i,\sigma_i^2)
\end{equation}
and $p(\sigma_i^2|\mu_i) = \textsf{Inv-Gamma}(\sigma_i^2|a,b)$.
In this case,
\begin{equation}
p(Y_i|\mu_i) \propto \textsf{Student}(\mu_i|m_i,s_i^2,\nu_i) ,
\end{equation}
where $(m_i,s_i^2,\nu_i)$ depend on $Y_i$ [and $(a,b)$].
Hyperparameters.
Sometimes it is convenient to express the prior via hyperparameters. In particular, suppose $\mu_1$ and $\mu_2$ are conditionally independent, conditional on the hyperparameter $\phi$:
\begin{equation}
p(\mu_1,\mu_2|\phi) = p(\mu_1|\phi)\,p(\mu_2|\phi) .
\end{equation}
The dependence between $\mu_1$ and $\mu_2$ follows from the joint variation due to $\phi$:
\begin{equation}
p(\mu_1,\mu_2) = \int p(\mu_1|\phi)\,p(\mu_2|\phi)\,p(\phi)\,d\phi .
\end{equation}
(The symmetric prior implied by the use of the hyperparameter may not be appropriate for your problem.)
Having introduced the hyperparameter, it may be convenient to delay integrating it out until a later stage of the analysis. Note that the joint posterior distribution can be expressed as
\begin{equation}
p(\mu_1,\mu_2,\phi|Y_1,Y_2) \propto p(Y_1|\mu_1)\,p(Y_2|\mu_2)\,p(\mu_1|\phi)\,p(\mu_2|\phi)\, p(\phi).
\end{equation}
A Gibbs sampler could be based on the complete set of full conditional posterior distributions:
\begin{align}
p(\mu_1|Y_1,\phi) &\propto p(Y_1|\mu_1)\,p(\mu_1|\phi) \\
p(\mu_2|Y_2,\phi) &\propto p(Y_2|\mu_2)\,p(\mu_2|\phi) \\
p(\phi|\mu_1,\mu_2) &\propto p(\mu_1|\phi)\,p(\mu_2|\phi)\,p(\phi) .
\end{align}
There is a sense in which the hyperparameter is the carrier of information between the entities.
Note that $\mu_1$ and $\mu_2$ are treated symmetrically here. In addition to learning about $\mu_2$ from $Y_1$, we also learn about $\mu_1$ from $Y_2$.