As others have pointed out, the nonassociativity of the octonions prevents one from constructing a group. For example, any subgroup of the octonions lives inside of a quaternion subalgebra. Having said that, the Clifford algebra $Cl(\mathbb{R}^7)$ has two inequivalent irreducible representations which are each as real vector spaces isomorphic to the octonions (i.e., they are eight-dimensional) and provided that we identify $\mathbb{R}^7$ with the imaginary octonions, the action of $Cl(\mathbb{R}^7)$ is given by left and right octonionic multiplications. This is analogous to what happens with $Cl(\mathbb{R}^3)$ substituting *octonion* for *quaternion* in what I said above. Now the Spin group $Spin(3)$ is the one-dimensional quaternionic unitary group and lives naturally inside $Cl(\mathbb{R}^3)$, so one could think of the group $Spin(7)$ as being the analogue of the one-dimensional octonionic unitary group.

By the same token, and given the low-dimensional isomorphisms
$$Spin(2,1) \cong SL(2,\mathbb{R})$$
$$Spin(3,1) \cong SL(2,\mathbb{C})$$
$$Spin(5,1) \cong SL(2,\mathbb{H})$$
one would be tempted to think of $Spin(9,1)$ as $SL(2,\mathbb{O})$, even though such a group as written does not exist.