Does the regularity estimate of the solution to a PDE problem give us the continuity from $H^{1}(\Omega)$ to $L^{2}(\Omega)$

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I'm studying the spectral theory of the Robin boundary PDE problem

\begin{align} & -\Delta u = f \text{ on } \Omega \\ & \dfrac{\partial u}{\partial \nu} + \alpha u= h \text{ on } \Gamma \end{align}

Where $\Omega$ is the domain, $\Gamma$ its border, $f \in L^{2}(\Omega)$

I must show that the operator $A : L^{2}(\Omega) \mapsto H^{1}(\Omega)$ such that $Af=u_f$ with $u_f$ being the solution to the problem is continuous (bounded).

Using Lax-Milgram theory and Friederichs identity, I have established that

$$\|u_f\|_{H^{1}(\Omega)} \leq C(\|f\|_{L^{2}(\Omega)}+\|h\|_{L^{2}(\Gamma)})$$

Is this enough to prove the continuity of $A$ ? Because I know that $B$ is bounded from $X$ to $Y$ iff $\|Bu\|_Y \leq \|u\|_X$

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I assume you have fixed a $h \in L^2(\Gamma)$, and you're asking whether $A \colon f \mapsto u_f$ is continuous. This can be easily verified using the estimate, as for $f_1, f_2 \in L^2(\Omega)$ we have \begin{align} -\Delta (u_{f_1}-u_{f_2}) = f_1-f_2 &\text{ on } \Omega \\ \dfrac{\partial }{\partial \nu}(u_{f_1}-u_{f_2}) + \alpha (u_{f_1}-u_{f_2})= h-h=0 &\text{ on } \Gamma. \end{align} Hence by the Lax-Milgram estimate we deduce that \begin{equation} \lVert u_{f_1} - u_{f_2}\rVert_{H^1(\Omega)} \leq C \lVert f_1 - f_2 \rVert_{L^2(\Omega)}, \end{equation} so $A$ is Lipschitz continuous.

It's worth pointing out that $A$ is not linear, but merely affine if $h \neq 0,$ so you can't apply the usual characterisation of continuity for linear operators.