Let $\sigma$ be the non-trivial element of the Galois group of the quadratic extension $L/K$. Let $\phi \colon E \to E_D$ be the isomorphism defined over $L$.

First, if $P \in E(\bar L)$ then $\sigma\bigl( \phi(P)\bigr) = - \phi\bigl( \sigma(P)\bigr)$. This is easy to check on the coordinates of $P=(x,y)$ by
$$\sigma\bigl(\phi(x,y)\bigr) = \sigma\bigl( x,y/\sqrt{D}\bigr) =\bigl( \sigma(x), \sigma(y)/\sigma(\sqrt{D})\bigr) = \bigl(\sigma(x), -\sigma(y)/\sqrt{D}\bigr) = - \bigl( \sigma(x), \sigma(y)/\sqrt{D}\bigr) = - \phi\bigl(\sigma(x),\sigma(y)\bigr)=-\phi\bigl(\sigma(x,y)\bigr).$$

Now $\phi$ induces a map $\phi_*\colon H^1(L,E)\to H^1(L,E_D)$. Let $\xi\in H^1(L,E)$ and let $g\in G_L$. Then write $*$ for the action of $\operatorname{Gal}(L/K)$ on these cohomology groups. We obtain
$$\begin{align*}\bigl(\sigma * \phi_*(\xi )\bigr)(g) &= \sigma\Bigl( \phi_*(\xi)(\sigma^{-1} g \sigma)\Bigr)\qquad\text{by def of $*$}\\
&=\sigma\Bigl( \phi\bigl(\xi(\sigma^{-1} g \sigma)\bigr)\Bigr)\qquad\text{by def of $\phi_*$}\\
&=-\phi\Bigl(\sigma\bigl(\xi(\sigma^{-1} g\sigma)\bigr)\Bigr)\quad\text{by the above}\\
&=-\phi \bigl( (\sigma * \xi)(g)\bigr)\qquad\text{by def of $*$} \\
&=\bigl(-\phi_*(\sigma * \xi)\bigr) (g) \qquad\text{by def of $\phi_*$}
\end{align*}
$$
As $\sigma*\phi_*(\xi)=-\phi_*(\sigma* \xi)$ holds for all $\xi\in H^1(L,E)$ it also holds for its subgroup $Ш(E/L)$. Now $(1+\sigma)*\bigl(\phi_*(\xi)\bigr) = \phi_*\bigl((1-\sigma)*\xi\bigr)$ shows that $\phi$ induces and isomorphism between $N\bigl( Ш(E_D/L)\bigr)$ and $(1-\sigma)\,Ш(E/L)$.