Imposing that an action should be conformally invariant has a subtle but important difference with respect to imposing that it should be diffeomorphism invariant. Let me try to emphasize the differences between Diffeomorphisms, Conformal transformations and Weyl transformations. I will also clarify how to impose the invariance of the action, namely I'll argue whether the fields or the coordinates must be changed.
Diffeomorphisms
The action on the coordinates is given by a general differentiable function with differentiable inverse: $x^\mu \to {x'}^\mu = {x'}^\mu(x)$. The action on fields with spin 0, 1 and 2 is the following
$$
\begin{align}
\phi(x) &\to \phi'(x') = \phi(x)\,,
\\
\partial_\mu\phi(x) &\to \partial'_\mu\phi'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\,,
\\
g_{\mu\nu}(x) &\to g'_{\mu\nu}(x') = \frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta}(x)\,.
\end{align}
$$
These transformations are just changes of variables, every theory is invariant under them.
Weyl transformations
Weyl transformations do not act on the coordinates but simply rescale the fields
$$
\begin{align}
\phi(x) &\to \tilde{\phi}(x) = \Omega^{-\Delta}(x)\phi(x)\,,
\\
\partial_\mu\phi(x) &\to \partial_\mu\tilde{\phi}(x) = \partial_\mu(\Omega^{-\Delta}(x)\phi(x))\,,
\\
g_{\mu\nu}(x) &\to \tilde{g}_{\mu\nu}(x) = \Omega^{2}(x)g_{\mu\nu}(x)\,.
\end{align}
$$
Conformal transformations
Conformal transformations are coordinate transformations that preserve the metric up to a scale factor, that is $x^\mu \to {x'}^\mu$ and
$$
\frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu} g_{\alpha\beta}(x)= \Omega^{-2}(x)g_{\mu\nu}(x)\,.\tag{1}
$$
In OP's example ${x'}^\mu = \lambda x^\mu$ and $\Omega(x) = \lambda$. Now, this is the key point: we don't want to impose that the theory is invariant under this change of variables, this is always true because it is a particular case of diffeomorphism invariance. We want instead to compare theories with the same metric. That is, we always imagine a conformal transformation as a diffeomorphism satisfying (1) followed by a Weyl rescaling that cancels the $\Omega$ factor. Therefore the transformation laws are
$$
\begin{align}
\phi(x) &\to \tilde{\phi}'(x') = \Omega^{-\Delta}(x)\phi(x)\,,
\\
\partial_\mu\phi(x) &\to \partial'_\mu\tilde{\phi}'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\,,
\\
g_{\mu\nu}(x) &\to \tilde{g}'_{\mu\nu}(x') = g_{\mu\nu}(x)\,.
\end{align}
$$
With the notation "tilde prime" I denote the composition of the two last transformations.
Invariance of the action
Consider a generic Lagrangian that depends on the field $\phi$, its first derivative and the metric tensor. The action then reads
$$
S = \int d^4x \, \mathcal{L}[g_{\mu\nu}(x),\phi(x),\partial_\mu\phi(x)]\,.
$$
The transformation acts only on the fields and leaves $x$ unchanged since it is a dummy variable (integrated over).
Proving diffeomorphism invariance amounts to requiring
$$
\begin{align}
S &= \int d^4x\, \mathcal{L}\left[g'_{\mu\nu}(x),\phi'(x),\partial'_\mu\phi'(x)\right] =
\\
&=\int d^4x'\, \mathcal{L}\left[g'_{\mu\nu}(x'),\phi'(x'),\partial'_\mu\phi'(x')\right] =
\\
&=\int d^4x\,|\Omega(x)|^4 \mathcal{L}\left[\Omega^{-2}(x)g_{\mu\nu}(x),\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\right]\,,
\end{align}
$$
in the second step I just renamed $x$ to $x'$, it's a trivial operation. In the last step I used the transformation properties of the measure and of the fields (See Di Francesco - Mathieu - Sénéchal for example).
A Weyl rescaling requires simply
$$
S = \int d^4x\, \mathcal{L}\left[\Omega^2(x)g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\partial_\mu(\Omega^{-\Delta}(x)\phi(x))\right]\,.
$$
Finally, the requirement for conformal invariance can be obtained by composing the two transformations above
$$
S = \int d^4x\, |\Omega(x)|^4\mathcal{L}\left[g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\right]\,.
$$
This is consistent with what OP says. Notice that since diffeomorphism invariance is always true, just proving Weyl invariance is enough. The converse is not obvious and it is object of current research. We could say that by imposing conformal invariance we are imposing Weyl invariance only by those $\Omega$'s that make $g_{\mu\nu}$ and $\Omega^2 g_{\mu\nu}$ diffeomorphic.