The answer is positive even in greater generality:
Suppose that $G$ is a locally compact group with (at most) countably many connected components, $M$ a locally compact Hausdorff topological space, $G\times M\to M$ a continuous action. Let $p\in M$ be a point whose $G$-orbit $Gp$ is closed in $M$. Then the orbit map induces a homeomorphism
$$
G/G_p \to Gp\subset M,
$$
where $Gp$ is equipped with the subspace topology. Here $G_p$ is the stabilizer of $p$ in $G$. This result is Theorem 2.13 in the book by D.Montgomery and L.Zippin "Topological Transformation Groups''.
Note that Theorem 2.13 is stated in the case of transitive group actions. However, in general, the action of $G$ on its orbit is transitive and the assumption that the orbit is closed implies that the orbit is a locally compact space (since we are assuming that $M$ is).
Note that for this to apply you have to assume that Lie groups are, say, 2nd countable (some people do not make this assumption), so that they have (at most) countably many components. Otherwise, the result is false.
For sake of completeness I am writing a solution based on the suggestion of Mike Miller:
Theorem: There are no topology $\tau$ and a compatible (finite-dimensional) smooth structure $A$ on $\text{Diff(M)}$ satisfying:
(1) The action of $\text{Diff(M)}$ on $M$ is continuous w.r.t $\tau$.
(2) $\text{Diff(M)}$ is a Lie group w.r.t $(\tau,A)$.
Proof:
We assume by contradiction there exist such a pair $(\tau,A)$.
Define $X_n = \{(p_1,...p_n)\in M^n |p_i \neq p_j \forall i \neq j \}$, and look at the map $\psi : \text{Diff(M)} \times X_n \rightarrow X_n$ ,
$\psi\left(\phi,(p_1,...p_n)\right)=(\phi(p_1),...\phi(p_n))$.
(1) It is easy to see that $X_n$ is an $n$-dimensional manifold. (It's an open subset of $M^n$).
(2) Continuity of the action of $\text{Diff(M)}$ on $M$ implies $\psi$ is continuous.
(3) n-transitivity of $\text{Diff(M)}$ implies the action $\psi$ (of $\text{Diff(M)}$ on $X_n$) is transitive.
Now fix some point $q=(q_1,...q_n)\in X_n$, and denote by $G_q = \{\phi \in \text{Diff(M)} | \phi(q)=q \} $ the stabilizer group. It is closed, hence by the closed subgroup theorem $G_q$ is an embedded Lie subgroup of $\text{Diff(M)}$.
So, The left coset space $\text{Diff(M)} / G_q$ is a topological manifold* of dimension $dim(\text{Diff(M)})-dim(G_q)$.
(It can also be given a unique smooth structure making the quotient map $\pi: \text{Diff(M)} \rightarrow \text{Diff(M)} / G_q$ a smooth submersion, but that is irrelevant for our discussion**).
Now , by easy proposition on continuous homogenous $G$-spaces (i.e topological spaces with a continuous transitive action of a topological group $G$) it follows that $\text{Diff(M)} / G_q$ and $X_n$ are homeomorphic.
In particular their dimension as topological manifolds are equal, hence:
$dim(\text{Diff(M)}) \ge dim(\text{Diff(M)})-dim(G_q) = dim(X_n)=n$ for every $n$ which is a contradiction.
*See Lee's book (Intro to smooth manifolds) Thm 21.17.
** If we assume the action of $\text{Diff(M)}$ on $M$ is smooth (not merely continuous), then we get that $\text{Diff(M)} / G_q$ and $X_n$ are diffeomorphic. (See Thm 21.18)
Best Answer
If I'm interpreting your (edited) question correctly, the answer is yes: For any element $g\in G$, $\theta_g:M\to M$ is smooth, since the action itself is smooth, and has a smooth inverse $\theta_{g^{-1}}:M\to M$, since $\theta_{g^{-1}}\circ\theta_g=\theta_{g^{-1}g}=\theta_e=\text{id}_M$. Thus each $\theta_g$ is a global diffeomorphism.