$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$
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 ) \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 $t>0$.