I first saw this problem on my 'droid and the picture was too small for me to make out what it was all about, so I drafted an answer based upon a somewhat different approach, viz. the derivative of the energy integral over $\Bbb R^3$ and not over the retrodgrade cone. But I thought it might be worth sharing here in any event. So here goes:
We start with the wave equation
$u_{tt} = \nabla^2 u = \nabla \cdot \nabla u; \tag 0$
as I wrote in my comment to the question, I think we need to take the energy as
$E = \dfrac{1}{2}\displaystyle \int_{\Bbb R^3} (u_t^2 + \nabla u \cdot \nabla u)\; d^3r; \tag 1$
then
$E_t = \dfrac{1}{2}\displaystyle \int_{\Bbb R^3} (2 u_t u_{tt} + 2\nabla u_t \cdot \nabla u)\; d^3r = \displaystyle \int_{\Bbb R^3} (u_t u_{tt} + \nabla u_t \cdot \nabla u)\; d^3r; \tag 2$
we note that
$\nabla \cdot (u_t \nabla u) = \nabla u_t \cdot \nabla u + u_t \nabla \cdot \nabla u, \tag 3$
which may be written
$\nabla \cdot (u_t \nabla u) - \nabla u_t \cdot \nabla u = u_t \nabla \cdot \nabla u. \tag 4$
We seek to integrate (4) over $\Bbb R^3$; to do so, we let $B(R)$ be the closed ball of radius $R$ in $\Bbb R^3$ centered at $(0, 0, 0)$; then
$\displaystyle \int _{B(R)} (\nabla \cdot (u_t \nabla u) - \nabla u_t \cdot \nabla u) \; d^3r = \int_{B(R)} u_t \nabla \cdot \nabla u \; d^3r; \tag 5$
by the divergence theorem we have
$\displaystyle \int _{B(R)} (\nabla \cdot (u_t \nabla u)) \; d^3r = \int_{S(R)} (u_t \nabla u) \cdot \vec n \; d\Sigma; \tag 6$
where $\vec n$ is the outward pointing unit vector field on $S(R)$, and $d\Sigma$ is the surface area element on $S(R)$ induced from $d^3r$ on $\Bbb R^3$. If we assume that $u_t \nabla u$ falls off sufficiently fast as $R \to \infty$, then we have
$\displaystyle \int_{\Bbb R^3} \nabla \cdot (u_t \nabla u) \; d^3r = \lim_{R \to \infty} \int_{B(R)} \nabla \cdot (u_t \nabla u) \; d^3r = \lim_{R \to \infty} \int_{S(R)} (u_t \nabla u) \cdot \vec n \; d\Sigma= 0; \tag 7$
we thus conclude that (5) implies
$\displaystyle \int_{\Bbb R^3} \nabla u_t \cdot \nabla u \; d^3r = \lim_{R \to \infty} \int_{B(R)} \nabla u_t \cdot \nabla u \; d^3r$
$= \displaystyle \lim_{R \to \infty} -\int_{B(R)} u_t \nabla \cdot \nabla u \; d^3r = -\int_{\Bbb R^3} u_t \nabla \cdot \nabla u \; d^3r; \tag 8$
it is now a straightforward matter to see that, with the aid of (8), (2) yields
$E_t = \displaystyle \int_{\Bbb R^3} (u_t u_{tt} - u_t \nabla \cdot \nabla u)\; d^3r = \int_{\Bbb R^3} u_t (u_{tt} - \nabla^2 u)\; d^3r = 0 \tag 9$
by virtue of (0). The conservation of the energy (1) is thus established.
We may use (9) to demonstrate the uniqueness of solutions to (0). For if $u$ and $v$ are solutions having the same initial conditions,
$u(x, 0) = v(x, 0) = f(x), \; u_t(x, 0) = v_t(x, 0) = g(x), \tag{10}$
where again, $f(x)$ and $g(x)$, as we have assumed of $u_t \nabla u$ in (7), decrease sufficiently rapidly for large $R = \Vert x \Vert$; then if we consider the function
$w(x, t) = u(x, t) - v(x, t), \tag{11}$
we see that $w(x, t)$ is also a solution to (0) by linearity, and its initial conditions are
$w(x, 0) = u(x, 0) - v(x, 0) = f(x) - f(x) = 0, \tag{12}$
$w_t(x, 0) = u_t(x, 0) - v_t(x, 0) = g(x) - g(x) = 0; \tag{13}$
it then follows that the energy $E[w]$ of the solution $w(x, t)$ satisfies
$E[w](0) = \dfrac{1}{2}\displaystyle \int_{\Bbb R^3} (w_t^2(x, 0) + \nabla w(x, 0) \cdot \nabla w(x, 0))\; d^3r = 0, \tag{14}$
and
$E_t[w](t) = 0, \; t \ge 0, \tag{15}$
which together imply
$E[w](t) = 0, \; t \ge 0; \tag{16}$
since both
$w_t^2(x, t), \nabla w(x, t) \cdot \nabla w(x, t) \ge 0, \tag{17}$
(14) and (16) force
$w_t(x, t) = 0, \; \nabla w(x, t) = 0 \; \forall x \in \Bbb R^3, t \ge 0; \tag{18}$
since the first derivatives of $w(x, t)$ all vanish it follows that $w(x, t)$ is constsnt for $x \in \Bbb R^3$, $t \ge 0$; since $w(x, 0) = 0$, we have
$w(x, t) = 0, \; x \in \Bbb R^3, t \ge 0; \tag{19}$
that
$u(x, t) = v(x, t), \; x \in \Bbb R^3, t \ge 0 \tag{20}$
now follows from (11), the definition of $w(x, t)$.
