Consider $$ u_t=u_{xx}+f(u) $$ with some non-linearity. Making the ansatz $u(x,t)=U(z), z=x-ct$ gives the ODE $$ U_{zz}+cU_z+f(U)=0. $$ Linearizing in $U$, gives the linear operator $$ L=\partial_{zz}+c\partial_z+f'(U). $$ This operator is not self-adjoint.
But with making the weighted ansatz $v(z)=e^{(c/2)z}u(z)$, we get the self-adjoint operator $$ \tilde{L}=\partial_{zz}-\frac{c^2}{4}-f'(U). $$
The kernel of $L$ is given by the span of $U'$, while the kernel of $\tilde{L}$ by the span of $e^{-(c/2)z}U'(z)$.
My question is whether $$ w\in(\textrm{ker}(\tilde{L}))^\perp~\Longleftrightarrow~w\in(\textrm{ker}(L))^\perp $$
The answer is no. I don't understand your deduction, but you say that $\ker L$ and $\ker\tilde L$ are one-dimensional spaces spanned by $g = U'$ and $\tilde g = e^{-(c/2)\cdot}U'$, respectively. These two functions are obviously linearly independent whenever $c\neq 0$. But for two closed subspaces $V$ and $W$ of a Hilbert space $H$ you have $V^\perp = W^\perp$ iff $V = W$. This is obviously not the case in your example (unless $c=0$).