You don't say what kind of a group $G$ is but I'm going to assume for simplicity that $G$ is finite. Then, yes, it follows from Artin-Wedderburn. The point is that once we know that $\mathbb{C}[G] \cong \prod \text{End}(V_i)$ as algebras we also know what modules look like in terms of the RHS: modules over a finite direct product canonically break up into a finite direct sum of modules over each factor, and modules over $\text{End}(V_i)$ are direct sums of copies of $V_i$. 

So if we consider the action of $\mathbb{C}[G]$ on a representation $V \cong \bigoplus_i n_i V_i$ then it follows immediately that the factors $\text{End}(V_j)$ where $n_j = 0$ act trivially, while the factors $\text{End}(V_i)$ where $n_i > 0$ act on each individual isotypic component and hence linear independently. 

It may be more conceptually satisfying to instead use here the <a href="https://en.wikipedia.org/wiki/Jacobson_density_theorem">Jacobson density theorem</a> or the <a href="https://qchu.wordpress.com/2012/11/11/the-double-commutant-theorem/">double commutant theorem</a>. 

**Edit:** Here are some details about a different way to organize the argument. We really just need to get clear on what the two-sided ideals of $\mathbb{C}[G]$ are. We have that

1. The matrix algebra $M_n(\mathbb{C})$ has no nontrivial two-sided ideals. This is true more generally of $M_n(D)$ for $D$ a simple ring, and we have more generally that the two-sided ideals of $M_n(R)$ are naturally in bijection with the two-sided ideals of $R$.

2. The two-sided ideals of a finite product $\prod R_i$ of rings are the tuples of two-sided ideals $\prod I_i$ of each factor.

3. Combining 1 and 2, the two-sided ideals of a finite product $\prod M_{n_i}(\mathbb{C})$ of matrix algebras are products $\prod I_i$ where each $I_i$ is either the entire matrix algebra $M_{n_i}(\mathbb{C})$ or zero. So the quotients $\prod M_{n_i}(\mathbb{C})$ are exactly given by deleting some of the factors. 

Now let $V$ be a complex representation of a finite group $G$ (not necessarily finite-dimensional). We get an induced homomorphism $\mathbb{C}[G] \to \text{End}(V)$ whose image, by 3, is obtained by deleting some of the factors in the Artin-Wedderburn decomposition $\mathbb{C}[G] \cong \prod_i M_{n_i}(\mathbb{C})$. A factor acts nontrivially on $V$ iff the corresponding irreducible appears in $V$ (this follows from thinking about the isotypic decomposition, I can give more details here if desired), so we conclude that the image of $\mathbb{C}[G]$ is the product of the $M_{n_i}(\mathbb{C})$ corresponding to all the irreducibles appearing in $V$ as desired.