Let $\mathcal{L}$ be the Lagrangian for usual QED with scalar, charged particles (with photons and electrons as well):
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m \right)\psi - \partial_{\mu}\phi^*\partial^{\mu}\phi-M^2\phi^*\phi$$
I was trying to show that, because of symmetries, that the Lagrangian above could be written in the same way if $\partial_{\mu} \to D_{\mu}$. However, I have to find $D_{\mu}$. The symmetry I am considering is such that
$$\psi \to \psi e^{i\Lambda(x)}$$ $$\phi \to \phi e^{i\Lambda(x)}$$ $$A_{\mu} \to A_{\mu}-\frac{1}{e}\partial_{\mu}\Lambda(x)$$
My attempt
Since $\psi$ and $\phi$ have no crossed terms between them, I figured that we should add a $\mathcal{L}_{int}$ (interactions) such that it would cancel the term
$$-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$$
that results from the fermionic part. Hence, $\mathcal{L}_{int} = -e\bar{\psi}\gamma^{\mu}A_{\mu}\psi$ seems to do the trick. However, the scalar particle bit does not seem to trivially simplify, as there are some remaining terms:
$$ieA_{\mu}\left[\left(\partial^{\mu}\phi^*\right)\phi-\phi^*\partial^{\mu}\phi\right]+e^2A_{\mu}A^{\mu}\phi^*\phi$$
This would give $D_{\mu} = \partial_{\mu}+ieA_{\mu}$.
Is my procedure correct? Is there a more intuitive way of solving this?
Edit: Corrected misuse of concepts as pointed in the comments.