$\newcommand{\Diff}{\operatorname{Diff}}$ First let $M$ be closed. Then $\Diff^\infty(M)$ is locally modelled on the space of smooth vector fields in $M$ which is a Frechet space. All Frechet spaces are homeomorphic. So this is reduced to the case you described. The same is true for $\Diff^k(M)$.

Let $M$ be open (non-compact without boundary). Then $\Diff^\infty(M)$ is open in $C^\infty(M,M)$ in the Whitney $C^\infty$ topology. But it is not locally contractible in this topology, and the arc connected component of the identity is contained in the group $\Diff^\infty_c(M)$ of diffeomorphisms which differ from the identity only on a compact set. The group $\Diff^\infty_c(M)$ is a regular Lie group locally modeled on the space of $\mathfrak X_c(M)$ smooth vector fields with compact support which is a nuclear (LF)-space. Similar for $\Diff^k(M)$. I do not remember whether $\mathfrak X_c(M)$ is a CW-complex.

See [Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001]
for lots of details on this, including the case of manifolds with boundary and with corners. See also section 41 of [here].

Edit: Vidit Nanda pointed out that the compact open $C^1$ or $C^\infty$ topology was asked for. This is not a good topology: In general it is not locally contractible, so no manifold modelled on topological vector spaces. Also $\Diff^k(M)$ is not open in $C^k(M,M)$ for any $k\ge 1$ if $M$ is not compact.

I once tried to describe a setting where $\Diff^\infty(M)$ with the compact $C^\infty$ topology would be a Lie group: smooth manifolds based on smooth curves instead of charts; but you need a lot of other structures like a geodesic structure, locally convex spaces as tangent spaces, and parallel transport. The resulting category of manifolds is monodially closed, and the the manifolds with finite dimensional (or even Banach) tangent spaces are exactly the usual ones. The topology is the final topology with respect to the smooth curves, and for $\Diff(M)$ it is indeed the compact $C^\infty$-topology. The theory is horrendibly complicated, and nobody ever used it. I have no idea whether this helps for the quest for CW-complexes. See:

• Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178. (pdf of part 1) (pdf of part 2)

Liviu Nicolaescu's answer seems to be the best for your question.

$\newcommand{\Diff}{\operatorname{Diff}}$ First let $M$ be closed. Then $\Diff^\infty(M)$ is locally modelled on the space of smooth vector fields in $M$ which is a Frechet space. All (infinite dimensional separable) Frechet spaces are homeomorphic. So this is reduced to the case you described. The same is true for $\Diff^k(M)$.

Let $M$ be open (non-compact without boundary). Then $\Diff^\infty(M)$ is open in $C^\infty(M,M)$ in the Whitney $C^\infty$ topology. But it is not locally contractible in this topology, and the arc connected component of the identity is contained in the group $\Diff^\infty_c(M)$ of diffeomorphisms which differ from the identity only on a compact set. The group $\Diff^\infty_c(M)$ is a regular Lie group locally modeled on the space of $\mathfrak X_c(M)$ smooth vector fields with compact support which is a nuclear (LF)-space. Similar for $\Diff^k(M)$. I do not remember whether $\mathfrak X_c(M)$ is a CW-complex.

See [Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001]
for lots of details on this, including the case of manifolds with boundary and with corners. See also section 41 of [here].

Edit: Vidit Nanda pointed out that the compact open $C^1$ or $C^\infty$ topology was asked for. This is not a good topology: In general it is not locally contractible, so no manifold modelled on topological vector spaces. Also $\Diff^k(M)$ is not open in $C^k(M,M)$ for any $k\ge 1$ if $M$ is not compact.

I once tried to describe a setting where $\Diff^\infty(M)$ with the compact $C^\infty$ topology would be a Lie group: smooth manifolds based on smooth curves instead of charts; but you need a lot of other structures like a geodesic structure, locally convex spaces as tangent spaces, and parallel transport. The resulting category of manifolds is monodially closed, and the the manifolds with finite dimensional (or even Banach) tangent spaces are exactly the usual ones. The topology is the final topology with respect to the smooth curves, and for $\Diff(M)$ it is indeed the compact $C^\infty$-topology. The theory is horrendibly complicated, and nobody ever used it. I have no idea whether this helps for the quest for CW-complexes. See:

• Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178. (pdf of part 1) (pdf of part 2)

Liviu Nicolaescu's answer seems to be the best for your question.

$\newcommand{\Diff}{\operatorname{Diff}}$ First let $M$ be closed. Then $\Diff^\infty(M)$ is locally modelled on the space of smooth vector fields in $M$ which is a Frechet space. All Frechet spaces are homeomorphic. So this is reduced to the case you described. The same is true for $\Diff^k(M)$.

Let $M$ be open (non-compact without boundary). Then $\Diff^\infty(M)$ is open in $C^\infty(M,M)$ in the Whitney $C^\infty$ topology. But it is not locally contractible in this topology, and the arc connected component of the identity is contained in the group $\Diff^\infty_c(M)$ of diffeomorphisms which differ from the identity only on a compact set. The group $\Diff^\infty_c(M)$ is a regular Lie group locally modeled on the space of $\mathfrak X_c(M)$ smooth vector fields with compact support which is a nuclear (LF)-space. Similar for $\Diff^k(M)$. I do not remember whether $\mathfrak X_c(M)$ is a CW-complex.

See [Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001]
for lots of details on this, including the case of manifolds with boundary and with corners. See also section 41 of [here].

$\newcommand{\Diff}{\operatorname{Diff}}$ First let $M$ be closed. Then $\Diff^\infty(M)$ is locally modelled on the space of smooth vector fields in $M$ which is a Frechet space. All Frechet spaces are homeomorphic. So this is reduced to the case you described. The same is true for $\Diff^k(M)$.

Let $M$ be open (non-compact without boundary). Then $\Diff^\infty(M)$ is open in $C^\infty(M,M)$ in the Whitney $C^\infty$ topology. But it is not locally contractible in this topology, and the arc connected component of the identity is contained in the group $\Diff^\infty_c(M)$ of diffeomorphisms which differ from the identity only on a compact set. The group $\Diff^\infty_c(M)$ is a regular Lie group locally modeled on the space of $\mathfrak X_c(M)$ smooth vector fields with compact support which is a nuclear (LF)-space. Similar for $\Diff^k(M)$. I do not remember whether $\mathfrak X_c(M)$ is a CW-complex.

See [Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001]
for lots of details on this, including the case of manifolds with boundary and with corners. See also section 41 of [here].

Edit: Vidit Nanda pointed out that the compact open $C^1$ or $C^\infty$ topology was asked for. This is not a good topology: In general it is not locally contractible, so no manifold modelled on topological vector spaces. Also $\Diff^k(M)$ is not open in $C^k(M,M)$ for any $k\ge 1$ if $M$ is not compact.

I once tried to describe a setting where $\Diff^\infty(M)$ with the compact $C^\infty$ topology would be a Lie group: smooth manifolds based on smooth curves instead of charts; but you need a lot of other structures like a geodesic structure, locally convex spaces as tangent spaces, and parallel transport. The resulting category of manifolds is monodially closed, and the the manifolds with finite dimensional (or even Banach) tangent spaces are exactly the usual ones. The topology is the final topology with respect to the smooth curves, and for $\Diff(M)$ it is indeed the compact $C^\infty$-topology. The theory is horrendibly complicated, and nobody ever used it. I have no idea whether this helps for the quest for CW-complexes. See:

• Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178. (pdf of part 1) (pdf of part 2)

Liviu Nicolaescu's answer seems to be the best for your question.