It will be Lipschitz continuous on any bounded rectangle since each of the component functions $f_1(x,y)=x \cos (y)$ and $f_2=x \sin (y)$ of the vector-valued function $f=(f_1,f_2)$ are $C^1$.
In particular, here we have $|\nabla f_1|^2=\cos^2(y)+x^2\sin^2(y)$ and $|\nabla f_2|^2=\sin^2(y)+x^2\cos^2(y)$. Hence $|\nabla f_1|,|\nabla f_2| \leq 1+x^2$. By the mean value theorem and the Cauchy-Schwarz inequality, we thus have
\begin{aligned}d\left( f(x,y), f(u,v) \right) &=\sqrt{\left(f_1(x,y)-f_1(u,v) \right)^2 +\left(f_2(x,y)-f_2(u,v) \right)^2 } \\& \leq |f_1(x,y)-f_1(u,v)|+|f_2(x,y)-f_2(u,v)|
\\& \leq 2\sqrt{(x-u)^2+(y-v)^2}+2\sqrt{(x-u)^2+(y-v)^2}
\\&=4\sqrt{(x-u)^2+(y-v)^2}
\\&=4d\left( (x,y), (u,v) \right) \quad \text { for all } (x,y), (u,v) \in [0,1] \times \mathbb{R} \end{aligned}
Note: Since our bound on each of $\nabla f_1$ and $\nabla f_2$ depends only on the first variable, $f$ is Lipschitz on every $G \subseteq T \times \mathbb{R}$ whenever $T \subset \mathbb{R}$ is bounded.
Notice $f$ can't be bi-Lipschitz on any set on which it fails to be injective. In particular, for every $K \geq 1$ and any $x \in [0,1]$ we will have
$$\frac{1}{K}d\left((x, 0), (x,2\pi)\right)=\frac{2\pi}{K}\geq 0=d\left(f(x, 0), f(x,2\pi)\right).$$
Note: $f$ will fail to be injective on any set $S \subseteq \mathbb{R}^2$ where there exist either $(x_1, y_1), (-x_1,y_2 ) \in S$ such that $y_1 \equiv \pi +y_2\pmod {2\pi}$ or $(x_1, y_1), (x_1,y_2 ) \in S$ such that $y_1 \equiv y_2 \pmod {2\pi}$.
Best Answer
You know that $x_1=x_2$, and you also know that $f(x_1,y_1)=f(x_2,y_2)$. So, this means that $f(x_1,y_1)=f(x_1,y_2)$. Then
$$e^{x_1}cos(y_1) = e^{x_1}cos(y_2)$$ and also $$e^{x_1}sin(y_1) = e^{x_1}sin(y_2)$$
Since $e^{x_1}>0$ for all values of $x_1$ from the definition of the exponential function, multiplying both sides of each equations above by $\frac{1}{e^{x_1}}$ yields
$$\begin{cases}cos(y_1) = cos(y_2) \\ sin(y_1) = sin(y_2)\end{cases}$$
But since we are considering $y\in ]0, 2\pi[$, if they have the same $cos$ and $sin$ simultaneously, then this implies that $$y_1 = y_2$$
So, $(x_1,y_1) = (x_2, y_2)$, then $f$ is one-to-one in $A$.
OBS: You can actually check the last statement by checking the quadrants in the unit circle. You know that: $$\begin{cases}cos(y_1) = cos(y_2) \\ sin(y_1) = sin(y_2)\end{cases}$$
And also that both $y_1,y_2\in]0,2\pi[$.Then, by analyzing all the cases:
If $sin(y_1)=sin(y_2) \geq 0$, if:
$cos(y_1)=cos(y_2) \geq 0$, then both $y_1,y_2\in]0,\frac{\pi}{2}]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
$cos(y_1)=cos(y_2) < 0$, then both $y_1,y_2\in]\frac{\pi}{2},\pi]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
If $sin(y_1)=sin(y_2) < 0$, if:
$cos(y_1)=cos(y_2) \geq 0$, then both $y_1,y_2\in[\frac{3\pi}{2},2\pi[$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.
$cos(y_1)=cos(y_2) < 0$, then both $y_1,y_2\in]\pi,\frac{3\pi}{2}]$, but both functions ($sin,cos$) are one-to-one in this interval, so this implies that $y_1=y_2$.