No nontrivial group $(G,\cdot)$ can be extended to a ring structure $(G,+,\cdot)$, because $G$ has no zero element with respect to the group operation $\cdot$. A zero element, also called an absorption element in a monoid, is an element $0\in G$ for which $g\cdot 0=0=0\cdot g$ for all $g\in G$. Cancelling $0$ from both sides of $0=0\cdot g$ would imply every $g\in G$ is the identity element, i.e. $G$ is trivial.

It's also not possible if you want $\mathrm{GL}_2(\mathbb{R})$ to be the monoid of nonzero elements in a ring. For if it was, that ring would be an associative four-dimensional division algebra (as every nonzero element has an inverse), which by the Frobenius theorem entails it must be the quaternions $\mathbb{H}$. But it cannot be the quaternions since $\mathbb{H}^\times\not\cong\mathrm{GL}_2(\mathbb{R})$ because $\mathbb{H}^\times$ only has two elements of order $2$ whereas $\mathrm{GL}_2(\mathbb{R})$ has at least four, $\mathrm{diag}(\pm1,\pm1)$.

(If topology were relevant we could proceed differently. Every quaternion has a polar form, a positive real times a unit quaternion, which entails $\mathbb{H}^\times\simeq \mathbb{R}\times\mathbb{S}^3$. On the other hand, $\mathrm{GL}_2(\mathbb{R})$ has two connected components corresponding to positive and negative determinant. Moreoever, $\mathrm{GL}_2^+(\mathbb{R})$ may be decomposed as a direct product of positive scalar multiples of $I_2$ times $\mathrm{SL}_2(\mathbb{R})$, and in tern $\mathrm{SL}_2(\mathbb{R})$ admits an Iwasawa decomposition, from which we may finally conclude that $\mathrm{GL}_2(\mathbb{R})\simeq\mathbb{R}^3\times \mathbb{S}^1\times\mathbb{S}^0$. Therefore $\mathbb{H}^\times$ and $\mathrm{GL}_2(\mathbb{R})$ cannot be homeomorphic, since they have different homotopy groups $\pi_0,\pi_1,\pi_3$.)