Before getting started, let us go back on some definitions:
Definition. Let $U$ be an open subset of $\mathbb{R}^n$ and let $f\colon U\rightarrow V\subseteq\mathbb{R}^n$ be a $C^1$-mapping, then $f$ is a $C^1$-diffeomorphism or globally invertible if and only if there exists $g\colon V\rightarrow U$ a $C^1$-mapping such that:
$$f\circ g=\textrm{id}_{V}\textrm{ and }g\circ f=\textrm{id}_U.$$
In other words, $f$ is a bijection whose inverse is smooth. This is not to be confused with:
Definition. The mapping $f$ is said to be a local diffeomorphism or locally invertible if and only if when retricted to an open subset of $U$, $f$ is a diffeomorphism onto its image.
Please notice that being globally invertible does not mean being locally invertible around any point.
As a reminder, let us state the inverse function theorem once again:
Theorem. Let $U$ be an open subset of $\mathbb{R}^n$, let $x$ be a point of $U$ and let $f\colon U\rightarrow\mathbb{R}^n$ be a $C^1$-mapping. Assume that $\mathrm{d}_xf\colon\mathbb{R}^n\rightarrow\mathbb{R}^n$ is an invertible linear map, then there exists $V$ an open neighbourhood of $x$ such that $f\colon V\rightarrow f(V)$ is a $C^1$-diffeomorphism.
In our case, for all $(x,y)\in\mathbb{R}^2$, the matrix of $\mathrm{d}_{(x,y)}f$ in the canonical basis of $\mathbb{R}^2$ is:
$$\begin{pmatrix}e^x\cos(y)&-e^x\sin(y)\\e^x\sin(y)&e^x\cos(y)\end{pmatrix}.$$
Its determinant is $e^{2x}$ which is nonzero for all $(x,y)$. Theorefore, according to the theorem, for all $(x,y)\in\mathbb{R}^2$, $f$ is a $C^1$-diffeomorphism in a neighborhood of $(x,y)$.
Assume by contradiction that there exists $g\colon\mathbb{R}^2\rightarrow\mathbb{R}^2$ a $C^1$-mapping such that:
$$g\circ f=\textrm{id}_{\mathbb{R}^2}.$$
Then, let $(X,Y)\in(\mathbb{R}^2)^2$ such that $f(X)=f(Y)$, then $g(f(X))=g(f(Y))$ i.e. $X=Y$ and $f$ is injective. However, $f$ is non injective since $f(0,0)=(1,0)=f(0,2\pi)$ and $(0,0)\neq (0,2\pi)$, a contradiction. Whence, $f$ is not a $C^1$-diffeomorphism.
Maybe it will give more insight to notice that if one identifies $\mathbb{R}^2$ with $\mathbb{C}$ through $(x,y)\mapsto x+iy$, then the considered mapping is the complex exponential, $z\mapsto e^z$ which is invertible on any horizontal strip of length at most $2\pi$.
Note that, since your domain is convex, you can join any two points $x, y$ by the line segment $c(t) = x + t(y-x)$.
Then
$$f(y)-f(x)=\int_0^1 df(x+t(y-x))(y-x)\, dt$$
Now mulitply this with $y-x$, use the fact that $df$ is positive and the integral linear, to conclude that this is not $=0$ if $y\neq x$.
Best Answer
Actually, this is not possible in $\mathbb{R}^n$ either.
Indeed, if you have any $\mathscr{C}^1$ injective function $f: \Omega \rightarrow \mathbb{R}^n$, then $f$ is open and a homeomorphism on its image (invariance of domain : https://en.m.wikipedia.org/wiki/Invariance_of_domain ).
From Sard’s theorem (https://en.m.wikipedia.org/wiki/Sard%27s_theorem ), the set of critical values has null measure in $\mathbb{R}^n$, thus has empty interior, thus the set of critical points has no interior as well.