It is well known that for a smooth manifold $M$, the de Rham cohomology group is defined by $$H_{dR}^k(M):=\frac{A^k(M)\cap \ker d}{A^k(M)\cap \text{im }d}.$$
Similarly, if we assume that $M$ being a compact smooth manifold, we can define a cohomology group by $$ H_{d^*}^k(M):=\frac{A^k(M)\cap \ker d^*}{A^k(M)\cap \text{im }d^*}, $$ since we have $d^*\circ d^*=0$.
Then what properties do we have for this cohomology group? Is there an isomorphism between $H_{dR}^k(M)$ and $H_{d^*}^k(M)$?
Indeed, $H^k_{d^*}(M) = H^k_{dR}(M)$. This holds for compact manifolds (with no boundary), since as mentioned in the post, the left is only well-defined on them.
This follows from the Hodge decomposition theorem, see https://en.wikipedia.org/wiki/Hodge_theory. In particular, it says that $$ A^k(M) \cong \operatorname{im} d_{k-1} \oplus \operatorname{im} d^*_{k+1} \oplus {\mathcal H}^k, $$ where ${\mathcal H}^k=\ker(dd^* + d^*d)$ is the space of harmonic forms. Harmonic forms are both closed and co-closed.
Then we see that $$ H^k_{dR}(M) = \frac{\ker d_{k}}{\operatorname{im} d_{k-1}} = \frac{\operatorname{im} d_{k-1}\oplus {\mathcal H}^k}{\operatorname{im} d_{k-1}}={\mathcal H}^k. $$
Similarly, $$ H^k_{d^*}(M) = \frac{\ker d_{k}^*}{\operatorname{im} d_{k+1}^*} = \frac{\operatorname{im} d_{k+1}^*\oplus {\mathcal H}^k}{\operatorname{im} d_{k+1}^*}={\mathcal H}^k. $$ Note here that no non-trivial element in $\operatorname{im} d_{k-1}$ can be $d^*$-closed, since $d^*(d\alpha)=0$ implies that $0=(d^*(d\alpha), \alpha)=\|d\alpha\|^2$.
Therefore, the two theories are the same.