Can a transcendent matrix have an algebraic spectrum?

Sure. The matrix could be nilpotent, hence all the eigenvalues would be $0$. It could also be upper or lower triangular with only elements of $K$ on the diagonal.

You can construct such a matrix easily enough. Let $K = \mathbf{C}$ and $K' = \mathbf{C}(t)$, then consider

$$ \begin{pmatrix} t & t+1 \\ 1-t & -t \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \\ \end{pmatrix} \begin{pmatrix} t & t+1 \\ 1-t & -t \\ \end{pmatrix}^{-1} = \begin{pmatrix} 2-t^2 & - t - t^2 \\ t^2 - t & t^2+1 \\ \end{pmatrix} $$

This matrix has eigenvalues $1$ and $2$ and eigenvectors $(t, 1-t)$ and $(t + 1, -t)$. There's nothing particularly special about these values, if you were to just pick any eigenvalues and eigenvectors, the resulting matrix would probably give you what you want.