Of course, there are many ways of metrizing the weak topology on $\mathcal M(\Omega)$ by using various tools of functional analysis. However, as it has already been pointed out by Dan, the most natural way is to use the transportation metric on the space of measures. [It is much more natural than the Prokhorov metric. I don't want to go into historical details here - they can be easily found elsewhere, but I insist that the transportation metric should really be related with the names of Kantorovich (in the first place) and his collaborator Rubinshtein]. Dan gives its dual definition in terms of Lipschitz functions, however its "transport definition" is actually more appropriate here. Let me remind it.
Given two probability measures $\mu_1,\mu_2$ on $\Omega$
$$
\overline d(\mu_1,\mu_2) = inf_M \int d(x_1,x_2) dM(x_1,x_2) \;,
$$
where $d$ is the original metric on $\Omega$, and the infimum (which is in fact attained) is taken over all probability measures $M$ on $\Omega\times\Omega$ whose marginals ($\equiv$ coordinate projections) are $\mu_1$ and $\mu_2$. One should think about such measures as "transportation plans" between distributions $\mu_1$ and $\mu_2$, while the integral in the RHS of the definition is the "cost" of the plan $M$.
It is obvious that the above definition makes sense not just for probability measures, but for any two positive measures $\mu_1,\mu_2$ with the same mass. Moreover, $\overline d(\mu_1,\mu_2)$ actually depends on the difference $\mu_1-\mu_2$ only, so that one can think about it as a "weak norm"
$$
|||\mu_1-\mu_2||| = \overline d(\mu_1,\mu_2)
$$
of the signed measure $\mu_1-\mu_2$ (clearly, it is homogeneous with respect to multiplication by scalars).
Let now $\mu=\mu_1-\mu_2$ be an arbitrary signed measure, where $\mu_1,\mu_2$ are the components of its Hahn decomposition. The only reason why the definition of the weak norm does not work in this situation is that the measure $\mu$ need not to be "balanced" in the sense that the total masses $\|\mu_1\|$ and $\|\mu_2\|$ need not be the same any more. However, this can be easily repaired in the following way: extend the original space $\Omega$
to a new metric space $\Omega'$ by adding to it an "ideal point" $o$ and putting $d(\omega,o)=1$ for any $\omega\in\Omega$. Then the measure
$$
\mu'=\mu - (\|\mu_1\|-\|\mu_2\|)\delta_o \;,
$$
where $\delta_o$ is the unit mass at the point $o$, is now balanced, so that $|||\mu'|||$ is well defined. Therefore, one can extend the definition of the weak norm $|||\cdot|||$ to arbitrary signed measures $\mu$ by putting
$$
|||\mu|||=|||\mu'||| \;.
$$
It is now easy to see that the distance $|||\mu_1-\mu_2|||$, where $\mu_1,\mu_2$ are two arbitrary signed measures, metrizes the weak topology on $\mathcal M(\Omega)$.
$\newcommand{\bR}{\mathbb{R}}$ For any Polish space $S$ (separable complete metric space) we denote by $M_1(S)$ the space of Borel probability measures on $S$.
The space $\newcommand{\eP}{\mathscr{P}}$ $\eP:=M_1(\bR)$ is a topological space with respect to the weak convergence. In fact, $\eP$ with this topology is a Polish space.
A random measure $\mu$ on $\bR$ is by definition a measure on $\eP$, i.e.,
$$\mu \in M_1\bigl(\;\eP\;\bigr).$$
The space $C_b(\bR)$ embeds in $C_b\bigl(\eP\bigr)$ $\newcommand{\bsE}{\boldsymbol{E}}$
$$ f\in C_b(\bR)\mapsto \widehat{f}\in C_b\bigl(\eP\bigr), \;\;\widehat{f}(\pi):=\bsE_\pi(f)=\int_{\bR} f(t) \pi(dt),\;\;\forall \pi\in \eP. $$
wher $\bsE_\pi$ denotes the expectation of a random variable with respect to the probability distribution $\pi$. Denote by $\newcommand{\eX}{\mathscr{X}}$ $\eX\subset C_b\bigl(M_1(\bR)\bigr)$ the image of this embedding.
In your question you are given $\newcommand{\bmu}{\boldsymbol{\mu}}$ a sequence $\bmu_n\in M_1(\eP)$ and another measure $\bmu$ with the property that
$$\lim_n\int_{\eP}F(\pi) \bmu_n(d\pi)=\int_{\eP}F(\pi)\bmu(d\pi),\;\;\forall F\in \eX $$
and you ask if the above holds for all $F\in C_b(\eP)$. Clearly this happens iff $\eX$ is dense in $C_b(\eP)$. It is not hard to see that is not the case. Fix a nontrivial, compactly supported continuous function $\rho:\bR\to[0,1]$. Then the bounded, continuous function
$$ Q_\rho:\eP\to\bR,\;\;Q_\rho(\pi)=\left(\int_{\bR}\rho(t) \pi(dt)\,\right)^2, $$
is not in the closure of $\eX$.
Best Answer
I believe that the answer is affirmative. Weak$^{\ast}$ topology on probability measures can be metrized (e.g. via Lévy–Prokhorov metric). When we deal with a Polish space $X$, then this metric is complete. Using this, I would like to conclude that the set of probability measures is sequentially closed; indeed, weak$^{\ast}$ convergent sequence of probability measures satisfies Cauchy condition, so it has to converge to some probability measure. Nevertheless, this space is definitely not weak$^{\ast}$ closed, as the example of $(\delta_{n}) \subset \ell_{\infty}^{\ast}$ shows.
I hope that it is correct.
EDIT The above argument is completely wrong: of course completely metrizable subset isn't necessarily closed, e.g. $(0,1) \subset \mathbb{R}$. Nevertheless, I found a paper by Dimitris Gatzouras, On weak convergence of probability measures in metric spaces, in which he claims that the set of separably supported Borel probability measures is sequentially closed. I haven't read it, so I can't tell if his argument is correct.