Perhaps it will be useful to know that in that case there is an isomorphism of the cohomology groups with compact support: $$ \pi_* : H^*_c(M\times\mathbb{R})\to H^{*-1}_c(M). $$ This is Proposition 4.7 in R. Bott, L. W. Tu, Differential forms in algebraic topology. Graduate Texts in Mathematics, 82. Springer-Verlag, New York-Berlin, 1982.
Regarding the Hodge decomposition we have the following result in Sobolev spaces.
Lemma ($L^p$-Hodge Decomposition). Let $M$ be a smooth, compact, $k$-dimensional oriented manifold without boundary and let $\Omega\subset M$ be an open subset. Then for any $p\in (1,\infty)$ and any $\ell$-form $\eta\in L^p(\bigwedge^\ell\Omega)$, $1\leq \ell\leq k$ there exist $\omega_1\in W^{1,p}(\bigwedge^{\ell-1}\Omega)$, $\omega_2\in W^{1,p}(\bigwedge^{\ell+1}\Omega)$ such that $$ \eta=d\omega_1+\delta\omega_2+h $$ where $h\in C^\infty(\bigwedge^\ell\Omega)$ is closed $dh=0$ and co-closed $\delta h=0$ and hence harmonic.
Theorem ($L^p$-Hodge Decomposition). Let $M$ be a smooth, compact, $k$-dimensional oriented manifold without boundary and let $\Omega\subset M$ be an open subset. Then for any $p\in (1,\infty)$ and any $\ell$-form $\eta\in L^p(\bigwedge^\ell\Omega)$, $1\leq \ell\leq k$ there exist $\omega_1\in W^{1,p}(\bigwedge^{\ell-1}\Omega)$, $\omega_2\in W^{1,p}(\bigwedge^{\ell+1}\Omega)$ such that $$ \eta=d\omega_1+\delta\omega_2+h $$ where $h\in C^\infty(\bigwedge^\ell\Omega)$ is closed $dh=0$ and co-closed $\delta h=0$ and hence harmonic.
I believe that if $\eta\in C^\infty$, then $\omega_1,\omega_2\in C^\infty$ so the classical case should be covered.
In the case when $\Omega=M$ the lemmatheorem is Proposition 6.5 in
C. Scott, $L^p$ theory of differential forms on manifolds. Trans. Amer. Math. Soc., 347 (1995), 2075-2096.
In the case of a general open set we simply extend $\eta$ to $L^p(\bigwedge^\ell M)$ by zero, apply the Hodge decomposition on $M$ and restrict all the resulting forms to $\Omega$.
Note that the above result applies to the manifold $M\times (0,1)$ since it can be isometrically embedded into $M\times S^1$ as an open set.
I know that this is not exactly the answer to the question, but I hope it could be helpful.