We prove that the assertion is true. This has already been proved in the literature, such as in [1, Corollary 8.6.3], and the following proof is also partially motivated by the proof therein.
1. Define $T : C_b(\mathbb{R}) \to \mathbb{R}$ by
$$ T(f) = \lim_{n\to\infty} \int_{\mathbb{R}} f \, \mathrm{d}\mu_n, \qquad \forall f \in C_b(\mathbb{R}). $$
This gives rise to a well-defined functional on $C_b(\mathbb{R})$ by the assumption. It is also clear that $T$ is a positive linear functional. Moreover, if $\| f \| := \sup \left| f \right| $ denotes the supremum norm, then
$$ \left| T (f) \right| \leq T(\left| f \right|) \leq \|f\| T(1), $$
and so, $T$ is continuous. Now by the Riesz-Markov-Kakutani representation theorem, there exists a unique Borel measure $\mu$ on $\mathbb{R}$ such that
$$ T(f) = \int_{\mathbb{R}} f \, \mathrm{d}\mu, \qquad \forall f \in C_c(\mathbb{R}). $$
In other words, we know that $\mu_n \to \mu$ in a vague topology. Moreover, it is clear that $\mu(\mathbb{R}) \leq T(1)$. Indeed, if $f \in C_c(\mathbb{R})$ and $0 \leq f \leq 1$, then
$$ \int_{\mathbb{R}} f \, \mathrm{d}\mu = T(f) \leq T(1), $$
and letting $ f \uparrow 1$ along a subsequence together with the monotone convergence theorem shows the desired inequality. This $\mu$ is certainly the very candidate for the weak limit of $\mu_n$.
2. To establish this convergence, however, we need the uniform tightness of $(\mu_n)_{n\geq 1}$. Considering that the tightness condition intuitively means "no mass escapes towards the infinity", this essentially amounts to showing that
$$\mu(\mathbb{R}) = T(1) = \lim_{n\to\infty} \mu_n(\mathbb{R}). $$
In order to show this, assume otherwise that $\mu(\mathbb{R}) < T(1)$. We will prove that this leads to a contradiction. Intuitively, we will construct a sequence of "bumps" $(\rho_j)_{j\geq 1}$ that moves along with the position of the mass escaping towards the infinity. This will then be used to derive a contradiction.
To this end, write $\epsilon = \frac{T(1) - \mu(\mathbb{R})}{2020} > 0$. Then we will recursively construct a seqeunce $(\rho_j)_{j\geq 1}$ in $C_c(\mathbb{R})$ and a subsequence $\mu_{n_j}$ as follows:
We will conveniently set $\chi_0 = 0$ and $n_0 = 0$.
Suppose that $j \in \{1, 2, \dots \}$, and $\chi_{j-1} \in C_c(\mathbb{R})$ with $0 \leq \chi_{j-1} \leq 1$ and $n_{j-1}$ has been chosen. Since
$$ \lim_{n\to\infty} \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu_n
= \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu
\leq \mu(\mathbb{R})
\qquad\text{and}\qquad
\lim_{n\to\infty} \mu_n(\mathbb{R}) = T(1), $$
we can find $n_j > n_{j-1}$ such that
$$ \int_{\mathbb{R}} \chi_{j-1} \, \mathrm{d}\mu_{n_j} < \mu(\mathbb{R}) + \epsilon \qquad \text{and} \qquad \mu_{n_j}(\mathbb{R}) > T(1) - \epsilon. $$
Also, choose $r > 0$ such that $\chi_{j-1}(x) = 0$ whenever $\left| x \right| > r$. Then we can find $\chi_j \in C_c(\mathbb{R})$ such that
$$ \mathbf{1}_{[-r-1,r+1]} \leq \chi_j \leq 1 \qquad \text{and} \qquad \int_{\mathbb{R}} \chi_{j} \, \mathrm{d}\mu_{n_j} > T(1) - 2\epsilon. $$
Finally, set $\rho_j = \chi_j - \chi_{j-1}$ for each $j \in \{ 1, 2, \dots \}$. Then
$$ \int_{\mathbb{R}} \rho_{j} \, \mathrm{d}\mu_{n_j} > T(1) - \mu(\mathbb{R}) - 3\epsilon, \qquad \forall j \geq 1. \tag{*} $$
Also, the construction ensures that $0 \leq \rho_j \leq 1$ for all $j $ and that for each given $x \in \mathbb{R}$ there is at most $2$ values of $j$ for which $\rho_j(x) > 0$. Altogether, for any bounded sequence $a = (a_j)_{j\geq 1} \in \ell^{\infty}$, the function
$$ f_a = \sum_{k \geq 1} a_k \rho_k $$
defines a bounded continuous function on $\mathbb{R}$. Then the assumption tells that
$$ \int_{\mathbb{R}} f_a \, \mathrm{d}\mu_{n_j}
= \sum_{k \geq 1} a_k \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu_{n_j} $$
converges as $j\to\infty$ for any $a \in \ell^{\infty}$. This shows that the sequence $c^{(j)} = (c^{(j)}_k)_{k\geq 1} \in \ell^1$ defined by $c^{(j)}_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu_{n_j} $ converges weakly as $j\to\infty$. Now it is well-known that this implies that $c^{(j)}$ converges in $\ell^1$. (See this posting, for instance.) Moreover, since
$$ \lim_{j\to\infty} c^{(j)}_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu, $$
the $\ell^1$-limit of $c^{(j)}$ is the sequence $c = (c_k)_{k\geq 1}$ defined by $c_k = \int_{\mathbb{R}} \rho_k \, \mathrm{d}\mu$. So we must have
$$ \lim_{j\to\infty} \left| c^{(j)}_j - c_j \right|
\leq \lim_{j\to\infty} \| c^{(j)} - c \|_{\ell^1}
= 0. $$
On the other hand, $\lim_{j\to\infty} c_j = 0$ and $\text{(*)}$ shows that
$$ \lim_{j\to\infty} \left| c^{(j)}_j - c_j \right|
\geq T(1) - \mu(\mathbb{R}) - 3\epsilon
> 0 $$
which is a a contradiction.
3. Now we are ready to prove the claim. Fix a sequence $\chi \in C_c(\mathbb{R})$ such that $0 \leq \chi(x) \leq 1$ for all $x \in \mathbb{R}$. Then for any $ f \in C_b(\mathbb{R})$ and $n$,
\begin{align*}
\left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right|
&\leq \left| \int f \chi \, \mathrm{d}\mu_n - \int f \chi \, \mathrm{d}\mu \right| \\
&\quad + \left| \int f(1 - \chi) \, \mathrm{d}\mu_n \right| + \left| \int f(1 - \chi) \, \mathrm{d}\mu \right| \\
&\leq \left| \int f \chi \, \mathrm{d}\mu_n - \int f \chi \, \mathrm{d}\mu \right| \\
&\quad + \|f\| \left( \mu_n(\mathbb{R}) - \int \chi \, \mathrm{d}\mu_n \right) + \|f\| \left( \mu(\mathbb{R}) - \int \chi \, \mathrm{d}\mu \right)
\end{align*}
Letting $n\to\infty$ and using the fact that $\mu(\mathbb{R}) = \lim_{n\to\infty} \mu_n(\mathbb{R})$,
\begin{align*}
\limsup_{n\to\infty} \left| \int f \, \mathrm{d}\mu_n - \int f \, \mathrm{d}\mu \right|
&\leq 2 \|f\| \left( \mu(\mathbb{R}) - \int \chi \, \mathrm{d}\mu \right)
\end{align*}
Since the left-hand side is independent of $\chi$, letting $\chi \uparrow 1$ along a subsequence proves the desired claim.
[1] Bogachev, Vladimir Igorevich. Measure Theory. Springer, 2007.
