Evaluated at $t=0$

$$\int _0^t u(x,t')\,dt'=0$$

whereas by assumption

$$(\int _0^t u(x,t')\,dt')_{t}=\int _0^t u_{t}(x,t')\,dt'+u(x,t)=g(x)\tag{5}$$

Also for each real $x$

$$ L(\int _0^t u(x,t')\,dt') = \int _0^t Lu(x,t')\,dt' + 2u_{t}(x,t) = 0\tag{6}$$

by formula $(1)$ and the equation in $(2)$. We thus recovered $(4)$ and uniqueness forces $v=\int _0^t u(x,t')\,dt'$. But then

$$\int _0^t u(x,t')\,dt' =\int _0^t \frac{1}{2}\bigl( g(x+ct') +g(x-ct')\bigr) dt' =

\frac{1}{2c} \bigl(\int _0^{x+ct}g(x')dx' -(-\int _{x-ct}^0 g(x')dx')\bigr)=\frac{1}{2c}\int_{x-ct}^{x+ct}g(x')\,dx'.\tag{7}$$

Someone else please do the Also.