How can we show this convolution identity for the exponential of measures on a group? Let $(G,\mathcal G)$ be a measurable group (think of a metric vector space, where the group operation is the addition, if you like) and $\mu,\nu$ be finite signed measures on $\mathcal M(G,\mathcal G)$.
Let $$\exp(\mu):=\sum_{k=0}^\infty\frac{\mu^{\ast k}}{k!},$$ where $\mu^{\ast k}$ denotes the $k$fold convolution.

I would like to show the identity $$\exp(\mu+\nu)=\exp(\mu)\ast\exp(\nu)\tag1.$$

I'm only able to show this under the assumption that $G$ is a normed $\mathbb R$-vector space, $\mathcal G=\mathcal B(G)$ and that $\mu$ and $\nu$ are tight$^1$, since it's easy to see that the characteristic functions of the left- and right-hand side of $(1)$ coincide and under the stated assumptions we know that the family $\{e^{{\rm i}\varphi}:\varphi\in G'\}$ is separating for the class of tight finite signed measures.

$^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called (uniformly) tight if for all $\varepsilon>0$, there is a compact $K\subseteq E$ with $\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon$. A finite signed measure $\mu$ is called tight if $\{\mu\}$ is tight.
 A: I assume that your group is commutative (otherwise I am not sure that the result is true). Recall that by definition of convolution $\mu * \nu$ is the unique measure such that for all mesurable function f :
$$ \int_{g \in G} f(g) ~ d(\mu * \nu)(x) = \iint_{g_1,g_2 \in G} f(g_1.g_2) ~ d\mu(g_1) \otimes d\nu(g_2).$$
Hence one has :
$$ \int_{g \in G} f(g) ~ dexp(\mu)(x) = \sum_{k=0}^{\infty} \frac{1}{k!} \iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ d\mu(g_1) \otimes \dots \otimes d\mu(g_k).$$
Let $f$ be a measurable function. We have to prove that :
$$ \int_{g \in G} f(g) ~ d(exp(\mu + \nu)(g) = \int_{g \in G} f(g) ~ d(exp(\mu)*exp(\nu)(g).$$
The RHS is equal to :
$$ \begin{eqnarray*} \int_{g \in G} f(g) ~ d(exp(\mu)*exp(\nu)(g)
&=& \iint_{x,y \in G} f(x.y) ~ dexp(\mu)(x) \otimes dexp(\nu)(y) \\
&=& \sum_{p=0}^{\infty} \sum_{q=0}^{\infty} \frac{1}{p!} \frac{1}{q!} \iiint_{x_1,\dots,x_p \in G} \iiint_{y_1,\dots,y_q \in G} f(x_1 \dots x_p.y_1 \dots y_q) ~  \\
& & \quad \quad \quad \quad  d\mu(x_1) \otimes \dots \otimes d\mu(x_p) \otimes d\mu(y_1) \otimes \dots \otimes d\mu(y_q) \\
\end{eqnarray*}$$
And the LHS is equal to :
$$ \begin{eqnarray*}
\int_{g \in G} f(g) ~ dexp(\mu + \nu)(g)
&=& \sum_{k=0}^{\infty} \frac{1}{k!} \iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ d(\mu+\nu)(g_1) \otimes \dots \otimes d(\mu+\nu)(g_k)
\end{eqnarray*}$$
Note that for every $k$, the product $d(\mu+\nu)(g_1) \otimes \dots \otimes d(\mu+\nu)(g_k)$ expands to :
$$ d(\mu+\nu)(g_1) \otimes \dots \otimes d(\mu+\nu)(g_k) = \sum_{I,J} \bigotimes_{i \in I} d\mu(g_i) \otimes \bigotimes_{j \in J} d\nu(g_j),$$
where $(I,J)$ runs over all partitions of $\{1,2,3,...,k\}$.
Hence the LHS is equal to :
$$ \begin{eqnarray*}
\int_{g \in G} f(g) ~ dexp(\mu + \nu)(g)
&=& \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{I,J} \iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ \bigotimes_{i \in I} d\mu(g_i) \otimes \bigotimes_{j \in J} d\nu(g_j)
\end{eqnarray*}.$$
Now for every $k$ and every partition $(I,J)$, since we assumed that the group is commutative, the integral $\iiint_{g_1,\dots,g_k \in G} f(g_1 \dots g_k) ~ \bigotimes_{i \in I} d\mu(g_i) \otimes \bigotimes_{j \in J} d\nu(g_j)$ depends only on the cardinal of $I$ and $J$ and is equal to
$$\iiint_{x_1,\dots,x_p,y_1,\dots,y_q \in G} f(x_1 \dots x_p.x_1 \dots x_p) ~ d\mu(x_i) \otimes \dots d\mu(x_p) \otimes d\nu(x_j) \dots \otimes d\nu(x_q),$$
where $p = \mathrm{Card}(I)$ and $p = \mathrm{Card}(J)$. And since for every pairs of integers $(p,q)$ such that $p+q=k$, there are $\frac{k!}{p!q!}$ partitions $(I,J)$ such that $p = \mathrm{Card}(I)$ and $p = \mathrm{Card}(J)$, you can see that LHS is equal to RHS.
