The functor of isomorphism classes of genus 1 curves over $R$ is not representable by a scheme or algebraic space since it's not an fppf sheaf. Descent fails for two reasons: there exist nontrivial twists, and there exist families of genus 1 curves $X \to S$ over a scheme $S$ such that $X$ is not a scheme but is a scheme étale locally on $S$.
Instead, we can consider the pseudofunctor $\mathcal{M}_1$ which associates to $R$ the groupoid of algebraic spaces smooth and proper over $R$ with fibers curves of genus 1. Then $\mathcal{M}_1$ is an algebraic stack which has a relatively explicit construction.
Let $\mathcal{E} \to \mathcal{M}_{1,1}$ be the universal elliptic curve. Then $\mathcal{M}_1$ is the quotient stack $$ \mathcal{M}_1 \cong [\mathcal{M}_{1,1}/\mathcal{E}] $$ where $\mathcal{E}$ acts trivially. The map $\mathcal{M}_1 \to \mathcal{M}_{1,1}$ sends a genus 1 curve to its Jacobian and the isomorphism is expressing the fact that every genus 1 curve is a torsor for its Jacobian.
$\mathcal{M}_{1,1}$ can also be expressed as the quotient of a scheme by a group using the Weierstrass equation so combining the two, we can write $\mathcal{M}_1$ as the quotient of a scheme by a group. Edit: this last claim is false, see Will Sawin's answer.
