Show that all complex Radon measures on a locally compact and $\sigma$-compact Hausdorff space is a Banach space

As is often the case, it is much easier to work with the following equivalent characterization of completeness of a normed vector space:

A normed vector space $(X, \|\cdot\|)$ is complete if and only if for each sequence $(x_n)_n$ in $X$ with $\sum_n \|x_n\| < \infty$, there is some $x \in X$ with $x = \sum_{n=1}^\infty x_n = \lim_{N\to\infty} \sum_{n=1}^N x_n$.

In your case, as you noted yourself, we know that $\mu(E) := \sum_{n=1}^\infty \mu_n (E)$ converges for every Borel set $E$.

Now, let $(E_n)_n$ be disjoint and let $\varepsilon >0$. There is $N$ with $\sum_{n=N}^\infty \|\mu_n\| < \varepsilon$. Hence, $$ \bigg|\sum_\ell \mu(E_\ell) - \mu(\biguplus_\ell E_\ell)\bigg| \leq \bigg|\sum_{n=1}^N \bigg(\sum_\ell \mu_n (E_\ell) - \mu_n(\biguplus_\ell E_\ell)\bigg)\bigg| + \sum_{n=N}^\infty \bigg|\sum_\ell \mu_n (E_\ell) \bigg| + \sum_{n=N}^\infty \bigg|\mu_n (\biguplus_\ell E_\ell)\bigg| < 2\varepsilon, $$ for every $\varepsilon > 0$.

Likwise, one can show regularity of $\mu$ and $\sum_{n=1}^N \mu_n \to \mu$ with convergence in $M(X)$.