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

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, 

(i) $\mu$ attaches masses $2/8,1/8,5/8$ to points $(0,0),(0,1),(1,0)$, respectively; 

(ii) $\hat\mu$ attaches masses $2/4,1/4,1/4$ to points $(0,0),(0,1),(1,0)$, respectively; 

(iii) $\nu$ attaches masses $1/4,1/4,2/4$ 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, 
$$(\mu * \nu)_1(\{0\})=\frac9{16}\ne\frac58=(\hat\mu * \nu)_1(\{0\}).$$