The answer is no. I know that for some people here, saying "It's false for $X = \ell^1$" would be a good enough hint, but I also know that this question originated on Math StackExchange, so I've included many of the details, making a rather long answer.

The spaces I use for a counterexample are $\newcommand{\R}{\mathbb{R}}X = \ell^1$, and $X_{*} = c_0$, using the usual pairing $\langle\mbox{-},\mbox{-}\rangle_1 : \ell^1 \times c_0 \rightarrow \R$ defined by:
$$
\langle \phi, a \rangle_1 = \sum_{n=0}^\infty \phi(n)\cdot a(n).
$$
I prefer a different notation from that used in the question, which I'll explain here. I'll use $\ell^1$ for that space considered as a Banach space, and $\ell^1_\sigma$ when equipped with the weak-* topology $\sigma(\ell^1,c_0)$. Then $L(\ell^1) = L(\ell^1,\ell^1)$ is the space of bounded/continuous linear maps from $\ell^1$ to itself in the norm topology, and $L(\ell^1_\sigma)$ the space of continuous linear maps $\ell^1_\sigma \rightarrow \ell^1_\sigma$.

The reason the answer is no is that $L(\ell^1_\sigma)$ is dense in $L(\ell^1)$, both in the $\sigma(L(\ell^1),\ell^1 \hat{\otimes} c_0)$-topology and the weak-* operator topology described in the question, but $L(\ell^1_\sigma)$ is a proper subset of $L(\ell^1)$. So $L(\ell^1_\sigma)$ is not closed in either of those topologies.

We'll begin with some definitions. For each $\newcommand{\N}{\mathbb{N}}n \in \N$ let's define $\delta_n : \N \rightarrow \R$ such that $\delta_n(m)$ is 1 if $n=m$, and $0$ otherwise. These functions belong to both $\ell^1$ and $c_0$ and are the usual Schauder bases of these spaces. Let's also define $e_n : \ell^1 \rightarrow \R$ by $e_n(\phi) = \phi(n)$. These functionals are weak-* continuous (*i.e.* $\sigma(\ell^1,c_0)$-continuous), essentially by definition.

We also define $s : \ell^1 \rightarrow \R$ by $s(\phi) = \sum_{i=0}^\infty \phi(i)$. This is norm continuous on $\ell^1$ but not weak-* continuous because $\delta_n \to 0$ in the weak-* topology, but $\phi(\delta_n) = 1 \not\to 0$. Therefore the rank 1 operator $T(\phi) = s(\phi) \cdot \delta_0$ is in $L(\ell^1)$ but not $L(\ell^1_\sigma)$. So all that remains is to show that $L(\ell^1_\sigma)$ is dense in $L(\ell^1)$. I'll give two proofs. I came up with the second one first, and it is simpler but requires more background knowledge. The first proof is more "bare hands" and is likely more what the OP was looking for.

In both cases, we actually show that the linear span of the rank one operators $T_{n,m}(\phi) = e_n(\phi) \cdot \delta_m \in L(\ell^1_\sigma)$ is dense in $L(\ell^1)$.

First proof:

The key statement in this proof is:

Given $f \in L(\ell^1)$, define:
$$
f_n = \sum_{i=0}^n \sum_{j = 0}^n f(\delta_i)(j) \cdot T_{ij}.
$$
This is a sequence of finite rank operators in $L(\ell^1_\sigma)$ converging to $f$ in the strong operator topology.

We prove this as follows. As part of the proof that $(\delta_n)$ is a Schauder basis for $\ell^1$, we know that for all $\phi \in \ell^1$, $\left(\sum_{i=0}^n \phi(i) \cdot \delta_i\right) \to \phi$. So for each $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \geq N$, $\| \phi - \sum_{i=0}^n \phi(i) \cdot \delta_i \| < \frac{\epsilon}{\| f \|}$. So for all $n \geq N$:

$$
\| f(\phi) - f_n(\phi) \| = \sum_{p = 0}^\infty |f(\phi)(p) - f_n(\phi)(p)|
$$
$$
= \sum_{p=0}^\infty \left| f(\phi)(p) - \sum_{i=0}^n \sum_{j=0}^n f(\delta_i)(j) \cdot e_i(\phi) \cdot \delta_j(p) \right|
$$
$$
= \sum_{p=0}^\infty \left| f(\phi)(p) - \sum_{i=0}^n f(\delta_i)(p) \cdot \phi(i) \right|
$$
$$
= \left\| f(\phi) - f\left(\sum_{i=0}^n \phi(i) \cdot \delta_i \right) \right\| \leq \|f\| \cdot \frac{\epsilon}{\|f\|} = \epsilon
$$
So $f_n \to f$ pointwise in $\ell^1$, *i.e.* in the strong operator topology. Since the weak-* topology is coarser than the norm topology, it also converges pointwise for that, so in the weak-* operator topology.

To prove that $f_n \to f$ in $\sigma(L(\ell^1),\ell^1 \hat{\otimes}c_0)$, we observe that by comparing the neighbourhood bases to each other, the weak-* operator topology and the topology $\sigma(L(\ell^1),\ell^1 \otimes c_0)$ agree (*i.e.* for the weak-* topology without completing the tensor product). Now, $\| f_n \| \leq \| f\|$, so we can use the following standard fact:

If $E$ is a Banach space, $D \subseteq E$ a dense subspace, then $\sigma(E^{*},E)$ and $\sigma(E^{*},D)$ agree on all norm bounded subsets of $E^{*}$.

I don't know a good reference for it in the above formulation, the closest being Schaefer's *Topological Vector Spaces* III.4.5 (taking $F = \R$ in that theorem). To finish, we apply it with $E^* = L(\ell^1)$, $E = \ell^1 \hat{\otimes} c_0$ and $D = \ell^1 \otimes c_0$, and the norm bounded set $\{ g \in L(\ell^1) \mid \| g \| \leq \| f \| \}$.

Second proof:

In this case, the key statement is

The set of maps $(T_{n,m})_{n,m \in \N}$ separates the points of $\ell^1 \hat{\otimes} c_0$ under the pairing with $L(\ell^1)$.

By a standard consequence of the Hahn-Banach theorem (see *e.g.* Schaefer's *Topological Vector Spaces* IV.1.3) this shows that the linear span of $(T_{n,m})_{n,m \in \N}$ is $\sigma(L(\ell^1),\ell^1 \hat{\otimes} c_0)$-dense. As the weak-* operator topology on $L(\ell^1)$ is coarser than this topology, this also proves density in that topology.

We use a theorem of Grothendieck to identify $\ell^1 \hat{\otimes} c_0$ with $\ell^1[c_0]$, *i.e.* the space of absolutely summable sequences in $c_0$. Under this isomorphism, the pairing of $L(\ell^1)$ with $\ell^1 \hat{\otimes} c_0$ is mapped to the following, where $f \in L(\ell^1)$ and $\Phi \in \ell^1[c_0]$:
$$
\langle f, \Phi \rangle_2 = \sum_{n = 0}^\infty \langle f(\delta_n), \Phi(n)\rangle_1
$$
If $\langle T_{n,m}, \Phi \rangle_2 = 0$ for all $n,m \in \N$, this means
$$
0 = \sum_{i=0}^\infty \langle T_{n,m}(\delta_i), \Phi(i) \rangle_1 = \sum_{i=0}^\infty \sum_{j=0}^\infty T_{n,m}(\delta_i)(j) \cdot \Phi(i)(j) = \sum_{i=0}^\infty \sum_{j=0}^\infty \delta_i(n)\cdot \delta_m(j) \cdot \Phi(i)(j)
$$
$$
= \Phi(n)(m)
$$
for all $n,m \in \N$, and therefore $\Phi = 0$. As discussed, this shows that the linear span of $(T_{n,m})_{n,m \in \N}$ is $\sigma(L(\ell^1),\ell^1[c_0])$-dense in $L(\ell^1)$ and so $L(\ell^1_\sigma)$ is also dense.