Yes, there is a generalization that covers this case, and much more general second order systems.  For example, you can consult L. P. Eisenhart's 1927 book *Non-Riemannian Geometry*, where he develops the geometry of paths (which is what you are asking about) along the line of his research on the subject with Veblen.  He has an effective test (in the form of the vanishing of a certain tensor) for when it is possible (locally) to map the all of the solution curves in $txy$-space to straight lines.  I don't have time to compute the tensor in your particular case right now, but it's not hard to do.

By the way, you should be aware that Laguerre's theorem is valid only locally.  You may not be able to define such a point transformation globally.  For example, the graphs in $ty$-space of the solutions to $y'' + y = 0$ cannot be globally mapped to straight lines because the graphs of solutions can cross each other multiple times.

**Additional Comment:**  I had a few minutes to look up the definition of the obstruction tensor in the case of the linear system given by the OP above and calculate it.   I got the following result:

Suppose one has a linear system of second order ODE of the form
$$
Z''(t) = A(t)\,Z'(t) + B(t)\,Z(t)
$$
where $Z(t) = (Z^i(t))$ is a column of height $n$ and $A(t)$ and $B(t)$ are $n$-by-$n$ matrices that are (say) smooth functions of $t$.  Then this system is point equivalent to $Z''(t) = 0$
if and only if 
$$
B_0(t) = \tfrac12 \, A'_0(t) - \tfrac14\,\mathrm{tr}\bigl(A(t)\bigr)\,A_0(t),
$$
where $A_0(t)$ and $B_0(t)$ are the trace-free parts of $A(t)$ and $B(t)$, respectively, i.e.,
$$
A_0(t) = A(t) - \tfrac1n\,\mathrm{tr}\bigl(A(t)\bigr)\,I_n\,
\quad\text{and}\quad
B_0(t) = B(t) - \tfrac1n\,\mathrm{tr}\bigl(B(t)\bigr)\,I_n\,.
$$

Note, by the way, that one can always reduce to the case $A(t)=0$ by point transformations, in which case the above condition simply becomes $B(t) = \lambda(t) I_n$ for some function $\lambda(t)$.