**False.**

Proof:

First, let us prove a short lemma:

For any $\displaystyle y_1, y_2 \in R$: $\displaystyle y_1 = y_2$ if and only if $\displaystyle y_1$ and $\displaystyle y_2$ have the same powers of of $\displaystyle y$ and same coefficients.

The proof is simple:

Let $\displaystyle a$ be the absolute value of the most negative power of $\displaystyle y$ in $\displaystyle y_1 + y_2$ ($\displaystyle a=0$ if there are no negative powers). For example, when $\displaystyle y_1 = y^{-2}+1+y$ and $\displaystyle y_2 = 3y$ we have $\displaystyle a=2$. Thus $\displaystyle y^a y_1, y^a y_2 \in \mathbb{C}[y]$, and therefore the powers and coefficients of $\displaystyle y^a y_1, y^a y_2$ and the same (since we assumed $\displaystyle y_1 = y_2$ for all $\displaystyle y$). The result follows.

Back to the original problem:

Let $\displaystyle n \in \mathbb{N}$ and $\displaystyle y_1, y_2 \cdots , y_n \in R$. As before, let $\displaystyle a$ be the absolute value of the most negative power of $\displaystyle y$ in $\displaystyle y_1 + y_1 + \cdots + y_n$, and $\displaystyle a=0$ if no negative powers exist. Thus, the most negative power in $\displaystyle y_1 z_1 + \cdots + y_n z_n$ for any $\displaystyle z_j \in \mathbb{C}[y]$ is no lower than $\displaystyle -a$. Therefore, the most negative power of $\displaystyle y$ in the set $\displaystyle y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$ is $\displaystyle -a$. From the lemma above and from the fact that the powers of $\displaystyle y$ in Laurent polynomials are not bounded, we conclude that there exist a Laurent polynomial $\displaystyle b$ for which $\displaystyle b \not\in y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$, so $\displaystyle y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y] \subset R$ for any choose of $\displaystyle n \in \mathbb{N}$ and $\displaystyle y_j \in \mathbb{C}[y]$. In other words, it is **false** that there exist an inetger $\displaystyle n \geq 1$ and $\displaystyle y, y_1, \cdots , y_n \in R$ such that $\displaystyle R=y_1 \mathbb{C}[y] + \cdots + y_n \mathbb{C}[y]$.