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The answer is no, in general it is not possible to express the measure $(\mu * \nu)_h$ in terms of the families $(\mu_h)_{h \in H}$ and $(\nu_h)_{h \in H}$ only. 

E.g., suppose that $G$ is the additive group $\R^2$, $N=\R\times\{0\}$, measures $\mu,\hat\mu,\nu$ are probability measures, 

$\bullet\quad$ $\mu$ attaches masses $\frac28,\frac18,\frac58$ to points $(0,0),(0,1),(1,0)$, respectively;
  
$\bullet\quad$ $\hat\mu$ attaches masses $\frac24,\frac14,\frac14$ to points $(0,0),(0,1),(1,0)$, respectively; 

$\bullet\quad$ $\nu$ attaches masses $\frac14,\frac14,\frac24$ to points $(0,0),(0,1),(1,0)$, respectively.  

Then $\mu_h=\hat\mu_h$ for $p_* \mu$-almost all $h \in H$ and, equivalently, for $p_* \hat\mu$-almost all $h \in H$. 

However, 
$$(\mu * \nu)_1(\{0\})=\frac9{16}\ne\frac58=(\hat\mu * \nu)_1(\{0\}).$$ 

---------------

**Added in response to the OP's modification of the original question by adding the semigroup requirement:**

Let $G$ and $H$ be as above. Let $(A_t)$, $(B_t)$, $(C_t)$, $(D_t)$ be independent standard Brownian motions.
For $t\ge0$, let 
\begin{equation}
	X_t:=\si_1 A_t,\quad Y_t:=\si_2(\rho A_t+\sqrt{1-\rho^2}B_t),
\end{equation}
where 
\begin{equation}
	\si_2:=\frac1{\sqrt{1-\rho^2}},\quad \si_1:=\rho\si_2=\frac\rho{\sqrt{1-\rho^2}}, 
\end{equation}
and $\rho\in(0,1)$. 

For each $t\ge0$, let $\mu_t$ and $\nu_t$ be the probability distributions of $(X_t,Y_t)$ and $(C_t,D_t)$, respectively, so that $\mu_t$ and $\nu_t$ are the bivariate normal distributions $N(0,0,\si_1^2 t,\si_2^2 t,\rho)$ and $N(0,0,t,t,0)$, and hence 
\begin{align*}
	\mu_t * \nu_t&=N\Big(0,0,(\si_1^2+1)t,(\si_2^2+1)t,\frac{\rho\si_1\si_2}{\sqrt{\si_1^2+1}\sqrt{\si_2^2+1}}\Big) \\ 
	&=N\Big(0,0,\frac t{1-\rho^2},\frac{(2-\rho^2)t}{1-\rho^2},
	\frac{\rho^2}{\sqrt{2-\rho^2}}\Big). 
\end{align*}
Moreover, obviously $(\mu_t)$ and $(\nu_t)$ are semigroups. 

Next, for each $h\in\R$ ($\R$ being identified with $H=\R\times\{0\}$), the measure $(\mu_t)_h$ is the conditional distribution of $Y_t$ given $X_t=h$, so that (cf. e.g. page 4 of [bivariate normal distribution][1] ) 
\begin{equation}
	(\mu_t)_h=N\Big(\rho\frac{\si_2}{\si_1}\,h,(1-\rho^2)\si_2^2 t\Big)
	=N(h,t),
\end{equation}
which **does not depend on $\rho$**. 
However, 
\begin{equation}
	(\mu_t *\nu_t)_h=N(\rho\sqrt{2-\rho^2}\, h,(2+\rho^2)t)
\end{equation}
obviously **does depend on $\rho$**.  

Thus, $(\mu_t *\nu_t)_h$ is not determined by $(\mu_t)_h$ and $(\nu_t)_h$, for any real $h$ and any $t>0$. 


  [1]: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwiL7c2-k5_aAhWprlQKHXT3C_cQFgg6MAE&url=http://athenasc.com/Bivariate-Normal.pdf&usg=AOvVaw1xNadZ0ODwcs2QCmbCduw9