It is said that "the symplectic geomtery is rather loose, like that of rubber sheet which may be stretched by different amounts in different directions and still retain its geomtery".
But the scaling transformation, in general, is not a symplectomorphism. For example, for $(x,\xi)\in \mathbb{R}^2$. Let $T(x,\xi)= (ax,b\xi)$ for some constants $a,b$ such that $ab\neq 1$. Let the symplectic form on $\mathbb{R}^2$ be given by \begin{equation} \sigma\Big((x,\xi),(y,\eta)\Big)=\xi y-\eta x. \end{equation} Clearly $\sigma\Big(T(x,\xi),T(y,\eta)\Big)=ab(\xi y-\eta x)\neq \sigma\Big((x,\xi),(y,\eta)\Big)$.
Can any one help me understand what has been said?
The symplectic form in $\Bbb R^2$ is the area form, so any scaling done in one direction must be undone in the other direction, as to preserve area. Such a scaling will be a symplectomorphism if and only if $b = 1/a$ (in view of your calculation, this might make more sense than explicitly saying that the condition is $ab=1$, but I suppose this is mainly psychological).
Maybe I should add the following generalization: if $V$ is any (finite-dimensional, say) vector space and $V^*$ is it's dual, then $V\oplus V^*$ has a symplectic structure given by $$\big((x,\xi), (y, \eta)\big) \mapsto``\begin{vmatrix} x & \xi \\ y & \eta\end{vmatrix}"= \eta(x) - \xi(y),$$and the transformation $$V\oplus V^* \ni (x,\xi) \mapsto (ax,b\xi) \in V\oplus V^*$$again will be a symplectomorphism if and only if $b = 1/a$.