It is not possible to define a normal derivative for all $u \in H^1(\Omega)$ which depends continuously on $u$.

The reason is that all $C_c^\infty(\Omega)$ is dense in $H^1(\Omega)$, but all $u \in C_c^\infty(\Omega)$ have zero normal derivative. If we would have a normal derivative depending continuously on $u \in H^1(\Omega)$, this implies that it is zero for all $u \in H^1(\Omega)$.

Your definition fails since $\Delta u \in H^1(\Omega)^*$ is meaningless for $u \in H^1(\Omega)$ (how do you define it?).

It is not possible to define a normal derivative for all $u \in H^1(\Omega)$ which depends continuously on $u$.

The reason is that all $C_c^\infty(\Omega)$ is dense in $H^1_0(\Omega)$, but all $u \in C_c^\infty(\Omega)$ have zero normal derivative. If we would have a normal derivative depending continuously on $u \in H^1(\Omega)$, this implies that it is zero for all $u \in H^1_0(\Omega)$ and this is not correct.

Your definition fails since $\Delta u \in H^1(\Omega)^*$ is meaningless for $u \in H^1(\Omega)$ (how do you define it?).