Let $X = h(Y)$. A first-order Taylor expansion around $E(Y) = \mu_y$ gives
$$X \approx h[\mu_y] + h'[\mu_y]\cdot [Y-\mu_y]$$
This easily leads to
$$\sigma^2_x \approx \big(h'[\mu_y]\big)^2\cdot \sigma^2_y$$
Since
$$\eta_{x,y} \equiv h'\cdot \frac Yh \implies \eta_{x,y}(\mu_y) = h'(\mu_y)\frac {\mu_y}{h(\mu_y)} \implies h'(\mu_y) = \eta_{x,y}(\mu_y) \cdot \frac {h(\mu_y)}{\mu_y}$$
Substituting,
$$\sigma^2_x \approx \left(\eta_{x,y}(\mu_y) \cdot \frac {h(\mu_y)}{\mu_y}\right)^2\cdot \sigma^2_y$$
Taking the square root we are led to
$$\frac {\sigma_x}{h(\mu_y)} \approx \eta_{x,y}(\mu_y) \cdot \frac {\sigma_y}{\mu_y} $$
The difference of the above from the expression in the question is that
a) The elasticity must be evaluated at the center of the Taylor expansion used
and
b) $h(\mu_y)$ is usually equal to $E(X)=E[h(Y)]=\mu_x$ only to a first approximation, due to Jensen's inequality.
Of course if the relation between $X$ and $Y$ is linear, and/or the elasticity is constant, things specialize.
In any case, it is valid as an approximation to write
$${\rm cv}_x \approx \eta_{x,y}(\mu_y) \cdot {\rm cv}_y $$
where "cv" stands for "coefficient of variation".
