Your suspicion is correct: such a $G'$ cannot exist. The idea is that a $G'$-structure determines a $G$-structure,
so if $M = \partial W$ as $G'$-manifolds, then $M = \partial W$ as $G$-manifolds.
In a bit more detail, suppose $\mathfrak U = \{(U, \phi_U\colon\mathbb R^n\to U)\}$ is an atlas for $M$ such that
$TM|_U\cong U\times\mathbb R^n$ for all charts $U$. If $U$ and $V$ are intersecting charts, there are transition functions $g_{UV} := d\phi_V^{-1}\circ d\phi_U$ from the description of $TM$ in $U$-coordinates to the description
in $V$-coordinates; they are invertible, so define maps $g_{UV}\colon U\cap V\to\mathrm{GL}_n(\mathbb R)$.
Moreover, the transition functions satisfy a cocycle condition on triple intersections $U\cap V\cap W$:
$$ g_{WU}\circ g_{VW}\circ g_{UV} = 1.$$
Now suppose $\rho\colon G\to\mathrm{GL}_n(\mathbb R)$ is a homomorphism of topological groups. There are a few ways
to define a $G$-structure on $M$; one hands-on definition is a lift of the transition functions across $\rho$. That
is, a $G$-structure is data of functions $\widetilde g_{UV}\colon U\cap V\to G$ for all $U,V\in\mathfrak U$ such
that $\rho\circ\widetilde g_{UV} = g_{UV}$ and the cocycle condition is satisfied: $\widetilde
g_{WU}\circ\widetilde g_{VW}\circ \widetilde g_{UV} = 1$ on all triple intersections. (We also have to specify that
$\widetilde g_{UU} = 1$.)
For example, a Riemannian metric determines an $\mathrm O_n$-structure; then an orientation is an
$\mathrm{SO}_n$-structure, and a spin structure is a $\mathrm{Spin}_n$-structure.
The point of all this is that if $\psi\colon G'\to G$ is a homomorphism of topological groups, $\rho\circ\psi\colon
G'\to\mathrm{GL}_n(\mathbb R)$ allows us to define $G'$-structures on manifolds. The key fact is that a
$G'$-structure $\{\widetilde g_{UV}\}$ determines a $G$-structure $\{\psi\circ\widetilde g_{UV}\}$, and this is
compatible with taking boundaries (a $G$-structure on $W$ induces a $G$-structure on $\partial W$ by restricting
the transition maps to the boundary). If
$[M] = 0$ in $\Omega_d^{G'}$, there is a compact $(d+1)$-dimensional $G'$-manifold $W$ such that $M = \partial W$
as $G'$-manifolds. Composing with $\psi$, we obtain $G$-structures on $M$ and $W$, and see that $M = \partial W$ as
$G$-manifolds, which contradicts the initial hypothesis that $[M]\ne 0$ in $\Omega_d^G$.