Thank you everyone for helping with this question. I would like to attempt to provide my own answer (which came to me after reading all the comments): Let $B_* (M)$ be rational, oriented bordism and $H_* (M)$ be the rational homology of $M$. I claim there is a map $$F:B_* (M) \rightarrow H_* (M)\otimes B_* (pt)$$ that is an isomorphism. The map $F$ sends $(P \rightarrow M) $ to $([P ] \otimes 1+1\otimes P')$ where $[P]\in H_* (M) $ represents the fundamental class of $P$ and $P' \in B_* (pt)$ is the bordism element represented by the abstract manifold $P$. This map is clearly an isomorphism when $M= pt$. The standard inductive argument over the number of cells implies this is an isomorphism in general. Note that no appeal to homotopy groups, spectral sequences or Thom spectra is being implictly used. We do use the computation of 0 dimensional bordism groups.

Is this correct?

Thank you everyone for helping with this question. I would like to attempt to provide my own answer (which came to me after reading all the comments): Let $B_* (M)$ be rational, oriented bordism and $H_* (M)$ be the rational homology of $M$. I claim there is a map $$F:B_* (M) \rightarrow H_* (M)\otimes B_* (pt)$$ that is an isomorphism. The map $F$ sends $(P \rightarrow M) $ to $([P ] \otimes 1+1\otimes P')$ where $[P]\in H_* (M) $ represents the fundamental class of $P$ and $P' \in B_* (pt)$ is the bordism element represented by the abstract manifold $P$. This map is clearly an isomorphism when $M= pt$. The standard inductive argument over the number of cells implies this is an isomorphism in general. Note that no appeal to homotopy groups, spectral sequences or Thom spectra is being implicitly used. We do use the computation of 0 dimensional bordism groups.

Is this correct?