Consider the entire function $$ A(z) = \pmatrix{e^z & 0\cr z & 1\cr}$$ An entire logarithm of $A(z)$ must have eigenvalues $z + 2\pi i n$ and $2 \pi i m$ for some (constant) integers $n$ and $m$, with eigenvectors $\pmatrix{(e^z-1)/z\cr 1\cr}$ and $\pmatrix{0\cr 1\cr}$ respectively (for $e^z \ne 1$). Such a matrix must be $$ \pmatrix{z + 2 \pi i n & 0\cr \frac{z^2 + 2 \pi i (n-m) z}{e^z - 1} & 2 \pi i m\cr}$$ But the $(2,1)$ matrix element has a pole at $z = 2\pi i j$ with $j \ne 0, m-n$, so this doesn't work.

Counterexample: Consider the entire function $$ A(z) = \pmatrix{e^z & 0\cr z & 1\cr}$$ An entire logarithm of $A(z)$ must have eigenvalues $z + 2\pi i n$ and $2 \pi i m$ for some (constant) integers $n$ and $m$, with eigenvectors $\pmatrix{(e^z-1)/z\cr 1\cr}$ and $\pmatrix{0\cr 1\cr}$ respectively (for $e^z \ne 1$). Such a matrix must be $$ \pmatrix{z + 2 \pi i n & 0\cr \frac{z^2 + 2 \pi i (n-m) z}{e^z - 1} & 2 \pi i m\cr}$$ But the $(2,1)$ matrix element has a pole at $z = 2\pi i j$ with $j \ne 0, m-n$, so this doesn't work.