Thom's famous paper "Quelques propriétés globales des variétés différentiable" gives enough information about the bordism ring of a point that one can, if I'm not mistaken, read off statements like this.
For unoriented bordism, he proves (Thm. II.10) that the classifying space $MO(k)$ has the $2k$-type of a product of mod 2 Eilenberg MacLane spaces. Hence the bordism group $\Omega_n^O(X)=\pi_{n+k}(MO(k)\wedge X)$ ($k \gg 0$) is isomorphic to $H_\ast(X; [H_\ast(X; \Omega^O_*(pt.))$.Omega^O_*(pt.))]_n$.
Presumably your question was about oriented bordism? In section 8 of his paper, Thom constructs the first few steps in a Postnikov tower for $MSO(k)$. But all that's relevant here is that $\Omega^{SO}_n(pt.)$ is $\mathbb{Z}$, $0$, $0$, $0$, $\mathbb{Z}$ for $n=0,1,2,3,4$, the generator isomorphism with $\mathbb{Z}$ in degree 4 being the signature. From the Atiyah-Hirzebruch spectral sequence it's then clear that $\Omega_n^{SO}(X) \cong H_n(X;\mathbb{Z})$ for $n=0, 1,2,3$. But $\Omega_4^{SO}(X)$ has an additional $\Omega_4^{SO}(pt.)$ \Omega_4^{SO}(pt.)=\mathbb{Z}$ summand which survives the spectral sequence, because it's the signature of the source manifold (a bordism invariant!).
The case of pairs $(X,A)$ can then be treated e.g. by Mayer-Vietoris.
