$X$ is certainly Calabi-Yau in the algebraic sense: the canonical line bundle is trivialized by the holomorphic volume form
$\Omega=\frac{dx}{x} \wedge \frac{dy}{y} \wedge \frac{du}{u}$
(this comes from the fact that $X$ is a conic bundle over $(\mathbb{C}^{*})^2$, degenerate over the curve $A=0$. The form $\Omega$ is constructed from the standard holomorphic volume form on $(\mathbb{C}^{*})^2$ and from the standard holomorphic volume on the $\mathbb{C}^{*}$ fiber).
In the compact Kähler case, this would be enough to insure the metric Calabi-Yau property, i.e. the existence of a Ricci flat metric, by Yau's theorem. But here, $X$ is not compact, so one has to be careful about question of completeness. My guess would be that in this particular example, there really exists a complete Ricci-flat metric but I don't know if/where it has been proved (if true).
In any case, what I have written is for any Laurent polynomial $A(x,y)$ and I don't think that the fact that $A$ comes from a knot should play a role (for this particular question).
EDIT (taking into account Vivek Shende's comment to the question): in my answer, I am assuming $A$ generic so that the curve $A=0$ is smooth and then $X$ is also smooth. But if $A=0$ is singular, as it is typical for the case coming from a knot (for a knot, there is always a factor $(y-1)$ if I remember correctly so the curve $A=0$ is always reducible), then $X=0$ will also be singular and so one has to be careful about what one means by Calabi-Yau).