Nice question (after it was rendered legible). Perhaps somewhat surprisingly, you cannot choose $y$ in this manner.
Multiplying the $x_i$ corresponds to adding their logarithms. By the central limit theorem, the distribution of $\log X_n$ will tend to a Gaussian with mean $n\mu$ and variance $n\sigma^2$, where $\mu$ and $\sigma^2$ are the mean and variance, respectively, of $\log x_i$. There are three possibilities:
If you choose $y$ such that $\mu\lt0$, eventually almost the entire distribution of $\log X_n$ will be concentrated at negative values, so $E[X_n]\lt1$.
If you choose $y$ such that $\mu\gt0$, eventually almost the entire distribution of $\log X_n$ will be concentrated at positive values, so $E[X_n]\gt1$.
If you choose $y$ such that $\mu=0$, the distribution of $\log X_n$ will remain concentrated symmetrically around $0$, but it will spread out and the upper part will blow up $E[X_n]$ while the lower part eventually contributes $0$. Thus $E[X_n]$ will increase without bound.