Let's try energy estimates (I'll assume constant coefficients, but variable coefficients shouldn't give too much trouble).

Since you are interested in uniqueness, we take two different solutions to your linear system with the same initial and boundary conditions and subtract them. Denote still by $u$ and $v$ the corresponding differences, now the inhomogeneity $f$ drops out.

Let us rewrite the transport equation for $v$ as
$$ \partial_t (e^{t} \nabla v) = e^{t} \nabla u $$
which we can integrate (and using that the initial data for the difference is 0) as
$$ \nabla v(t) = \nabla \int_0^t e^{s-t} u(s) ds $$
hence
$$ L_2 v(t) = L_2 \int_0^t e^{s-t} u(s) ds $$
Plugging this into the first equation, integrating by parts and assuming sensible boundary conditions, you should have the differential form of the energy identity looking like

$$ \frac{d}{dt} \int \frac12 |\partial_t u|^2 + \frac12 A_1(\nabla u, \nabla u) + A_2( \nabla \int_0^t e^{s-t} u(s) ds, \nabla u) dx = \int A_2(\nabla u, \nabla u) - A_2(\nabla \int_0^t e^{s-t} u(s) ds, \nabla u) dx $$

Denote by $E(t) = \sup_{s \in [0,t]} \int |\partial_t u(s)|^2 + A_1 (\nabla u(s), \nabla u(s)) dx$

We can estimate, depending on how $A_1$ compares with $A_2$ (in terms of ellipticity)
$$ |\int A_2( \nabla \int_0^t e^{s-t} u(s) ds, \nabla u) dx| \leq C t E(t) $$

So we have that the energy estimates imply

$$ E(t) - Ct E(t) \leq C \int_0^t E(s) + C s E(s) ~ds $$

For $t > 0$ sufficiently small we can absorb $Ct E(t)$ on the left, and get

$$ E(t) \leq C' \int_0^t E(s) + C s E(s) ~ ds $$

and by Gronwall's lemma, since $E(0) = 0$, we have $E(t) \equiv 0$ (for all $t$ sufficiently small, with smallness depending only on the ellipticity of $A_1$ and $A_2$, so we can iterate and get that this holds for all times).

This implies that $u\equiv 0$, which implies that $e^t \nabla v \equiv 0$ so that $v$ is constant on every given time. If you have sane boundary conditions this also implie that $v \equiv 0$.

For the case that the equation satisfied by $v$ is $\triangle ( v_t + v - u) = 0$, you will need to do additional elliptic estimates, but I think the basic idea should still work.