Pre-measurement
Let $\left| \psi \right>$ be a pure state:
$$ \left| \psi \right> = a \left| 0 \right> + b \left| 1 \right> $$
The density matrix of $\left| \psi \right>$ is:
$$ \hat{\rho}=\pmatrix{ aa^* & ab^* \\ ba^* & bb^* } $$
The entropy $S=\text{Tr} \hat{\rho} \ln \hat{\rho}$ is:
$$ S=\text{Tr} \pmatrix{ aa^* & ab^* \\ ba^* & bb^* } \ln \pmatrix{ aa^* & ab^* \\ ba^* & bb^* } $$
It is solved as follows:
$$ \begin{align} S&=\text{Tr} U^\dagger \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } U U^\dagger \ln \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } U\\ &=\text{Tr} \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } \ln \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } \\ &=(aa^* + bb^*)\ln(aa^* + bb^*) \end{align} $$
Then since $aa^* + bb^*=1$, $S=0$. (We have taken the convention that $0\ln0=0$.)
Post-measurement
Consider that after a measurement, the density matrix expresses a mixture of states (lets call it $\hat{m}$):
$$ \hat{m}=\pmatrix{ aa^* & 0 \\ 0 & bb^* } $$
The entropy of $\hat{m}$ is:
$$ \begin{align} S&=\text{Tr} \pmatrix{ aa^* & 0 \\ 0 & bb^* } \ln \pmatrix{ aa^* & 0 \\ 0 & bb^* }\\ &=aa^* \ln aa^*+ bb^*\ln bb^* \end{align} $$
which is higher than $0$ (as expected from a measurement)
This entropy is the same as the Shannon entropy: it quantifies the information gained by selecting of one element $\left| 0 \right>$ or $\left| 1 \right>$ from the set $\{ \left| 0 \right>,\left| 1 \right> \}$ according to the probability $\rho(\left| 0 \right>)=aa^*$ and $\rho(\left| 1 \right>)=bb^*$:
$$ \begin{align} H&=-\sum_{i=0}^n \left< i \middle| i \right> \ln \left< i \middle| i \right>\\ &=aa^* \ln aa^*+ bb^*\ln bb^* \end{align} $$
So far so good
Problem
Now, here is the problem I am having.
After applying a unitary transformation $U$ to $\left| \psi \right>$, the Von Neumann entropy of both the state ($S(\hat{\rho})=S(U^\dagger \hat{\rho} U)$) and the mixture ($S(\hat{m})=S(U^\dagger \hat{m} U)$) remains the same, but the Shannon entropy of the measurement changes.
Applying $U$ to $\left| \psi \right>$, we get:
$$ \begin{align} \left| \psi' \right> &= \underbrace{\pmatrix{a & b \\ -b^* & a^*}}_{U} \left| \psi \right>\\ &= \pmatrix{a & b \\ -b^* & a^*} \pmatrix{a \\ b}\\ &= aa+bb \left| 0 \right> + -ab^* + ba^* \left| 1 \right> \end{align} $$
The Shannon entropy of this measurement is :
$$ H = (aa+bb)\ln(aa+bb) + (-ab^* + ba^*)\ln (-ab^* + ba^*) $$
Why the departure between the two definitions of entropy? Which is lying?
To me, it looks like it is the Von Neumann entropy that is lying: implicit in the trace operator, the entropy is defined in reference to the eigenbasis. But, this will only be true if the future upcoming measurement is performed against the eigenbasis. If the future measurement is performed against another basis, the entropy will not be the Von Neumann entropy. The Shannon entropy, however, is always correct.