After translating the graph of y = sin(x) $\frac{\pi}{6}$ units to the left, we get the graph of y = sin($x + \frac{\pi}{6}$).
Then, changing each point's abscissa to $\frac{1}{\omega}$ times the original, we obtain the graph of f(x) = sin($\omega x + \frac{\pi}{6}$).
In the interval (0, $\frac{\pi}{2}$), $\omega x + \frac{\pi}{6} \in (\frac{\pi}{6}, \frac{\omega \pi}{2} + \frac{\pi}{6} )$.
If the function y = f(x) has only one zero in the interval (0, $\frac{\pi}{2}$), then $\frac{\omega \pi}{2} + \frac{\pi}{6} \in (\pi, 2\pi]$, hence ω ∈ $(\frac{5}{3}, \frac{11}{3}]$.

Thus, the answer is: $\boxed{B: (\frac{5}{3}, \frac{11}{3}]}$.

This solution uses the properties of the graph transformation rules for the function y = Asin($\omega x + \varphi$) and the zeroes of the sine function. The problem primarily assesses understanding of the graph transformation rules for functions of the form y = Asin($\omega x + \varphi$) and the zeroes of the sine function, making it a fundamental question.