A function is a diffeomorphism if the function is differentiable with a differentiable inverse. That is $ f:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m} $ is a diffeomorphism if f is differentiable, $ f^{-1} $ exists, and $ f^{-1} $ is differentiable.
$ f(x,y)=(e^{x}cos(y),e^{x}sin(y)) $ is not invertible on its domain, $ \mathbb{R}^{2} $. You have the proof in your question: $ f(x,y+2\pi k)=f(x,y) $.
Let me be try to be precise...
A function is invertible if it is injective and surjective.
A function is injective if '$ f(a)=f(b) $ implies $ a=b $' or equivalently '$ a\neq b $ implies $f(a)\neq f(b)$'. [Here $a,b$ are points in $\mathbb{R}^{n}$ or the Domain, and $f(a),f(b)$ are points in $\mathbb{R}^{m}$ or the Range - in this case both $\mathbb{R}^{2}$]
As an example, consider the following two points in $\mathbb{R}^{2}$ : $a=(0,0)$ and $b=(0,2\pi )$. We can see that $a\neq b$, but $f(a)=f(0,0)=(1,0)=f(0,2\pi )=f(b)$. This violates the definition of injectivity. It follows that $f$ is not invertible.
Since it is required that our function be invertible in order to be a diffeomorphism, the function above (although differentiable) is not a diffeomorphism.
Here are some links on invertible functions. I hope this helps.
http://en.wikipedia.org/wiki/Inverse_function
http://gowers.wordpress.com/2011/10/11/injections-surjections-and-all-that/