For simplicity, consider solutions where $ u$ does not depend on $x, y$:
$A u_t + B(t) u = 0$. If $y^T A = 0$, that says $y^T B(t) u = 0$, so $u$ is
restricted to belong to a certain (possibly $t$-dependent) subspace.
Thus for your example $$A = \pmatrix{1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & -1\cr}$$
$(B(t) u)_2 = (B(t) u)_3 = 0$. If the appropriate $2 \times 2$ submatrix of $B(t)$ is invertible, this lets you express $u_2$ and $u_3$ in terms of $u_1$ and $u_4$, and you get a periodic linear system for $u_1$ and $u_4$.
The solutions
are usually not periodic in $t$. Rather, the linear operator $u(0) \to u(2\pi)$ will have eigenvalues $\lambda$ corresponding to solutions where $u(2\pi) = \lambda u(0)$ (see Floquet theory).
Somewhat more generally, solutions of the form $u(x,y,t) = \exp(\alpha x + \beta y) v(t)$ lead to the same type of system (with $\alpha \cos(t) + \beta \sin(t)$ added to $B$).
EDIT: If the submatrix of $B(t)$ is not invertible for some $t$, you may find that some or all of the nontrivial solutions have singularities at those $t$.