First off, some facts. Every isometry fixing the origin is an orthogonal linear transformation of space, the group of all such linear operators is called $O(V)$. The ones which preserve the orientation of space (equivalently, are path-connected to the identity map) form a subgroup called the special orthogonal group $SO(V)$, which has index two.

A hyperplane is a codimension one subspace, or equivalently the orthogonal complement of a 1D line, and a reflection is an isometry which acts as negative one on a line and as the identity on its orthogonal complement (a hyperplane). Reflections are in $O(V)$ but not $SO(V)$, and every isometry is a composition of reflections. In particular, a composition of reflections across the complements of lines $\ell_1$ and $\ell_2$ will be a rotation in the 2D plane $\mathrm{span}(\ell_1,\ell_2)$ by an angle twice that of the angle between $\ell_1$ and $\ell_2$.

A plane rotation is a map which acts as a usual rotation on a 2D subspace and acts as the identity map on the plane's orthogonal complement. Plane rotations are elements of $SO(V)$. Two plane rotations with respect to orthogonal planes will commute with each other. Every rotation, i.e. every element of $SO(V)$, is a product of plane rotations with respect to a collection of pairwise orthogonal 2D planes. If $R$ is a plane rotation with respect to the (oriented) plane $\Pi$ by an angle of $\theta$, and $S$ is any other isometry, then $SRS^{-1}$ will be a plane rotation with respect to the oriented plane $S\Pi$ by an angle of $\theta$.

In a normed division algebra $\mathbb{K}$, there is a multiplicative norm $|\cdot|$ (i.e. it satisfies $|xy|=|x||y|$), and it induces an inner product $\langle-,-\rangle$ satisfying $|x|^2=\langle x,y\rangle$. The imaginary elements are those that are perpendicular to $1$ with respect to this inner product. If $u,v$ are orthogonal imaginaries, then they anticommute, i.e. $uv=-vu$

More generally, for any imaginaries $u$ and $v$, there is a formula $uv=-\langle u,v\rangle+u\times v$. For quaternions, $u\times v$ is the usual cross product, but for octonions it is a much less symmetrical seven-dimensional cross product (which has symmetry group $G_2\subset SO(7)$).

The division algebra $\mathbb{K}$ will also be alternative. This means every pair of elements generates an associative subalgebra, even if $\mathbb{K}$ itself isn't associative. Concretely this means the algebraic expressions $aab$, $aba$ and $baa$ are unambiguous without parenthesization for all $a,b$.

The square roots of negative one are precisely the unit pure imaginaries $u$. Since $\mathbb{K}$ is alternative, $u(ux)=(uu)x=-x$ makes sense, and thus $\mathbb{O}$ becomes a left and right four-dimensional complex vector space over $\mathbb{R}[u]$ with scalar multiplication from the left and right. The two sides don't necessarily match: multiplying anything in $\mathrm{span}(1,u)$ on either side does the same thing, but $uv=-vu$ for all $v$ in the orthogonal complement of $\mathrm{span}(1,u)$.

In the quaternions $\mathbb{H}$, given any unit imaginary $u$ we can form an oriented orthonormal basis $1,u,v,w$ (satisfying the same relations as $1,i,j,k$). Then multiplying by $e^{\theta u}=\cos(\theta)+\sin(\theta)u$ on the left will rotate $\mathrm{span}(1,u)$ and $\mathrm{span}(v,w)$ each by $\theta$, whereas right multiplication will rotate $\mathrm{span}(1,u)$ and $\mathrm{span}(v,w)$ by $\theta$ and $-\theta$ respectively. If we combine the two actions using a half angle, we find that $x\mapsto e^{(\theta/2)u}xe^{-(\theta/2)u}$ acts as the identity on $\mathrm{span}(1,u)$ and a rotation by $\theta$ on $\mathrm{span}(v,w)$. If we restrict this conjugation action to the 3D subspace of imaginary quaternions, this is rotation around the oriented axis $\mathbb{R}u$ by an angle of $\theta$.

The unit quaternions are closed under multiplication and form a group called $\mathrm{Sp}(1)$ or $S^3$ (some abusively call it $\mathrm{SU}(2)$ which I dislike). Using this conjugation action, we get a group homomorphism $\mathrm{Sp}(1)\to\mathrm{SO}(3)$ with kernel $\pm1$.

The same thing happens in $\mathbb{O}$, except now $e^{(\theta/2)u}xe^{-(\theta/2)u}$ will be a rotation in three different planes within the 7D space of imaginary octonions. I don't see any obvious description of which three planes (indeed, the three planes are not an invariant - one could use different sets of three planes to describe the same rotation, such is the ambiguity of isoclinic rotations). I assume they generate $SO(7)$ but it's not clear to me how. One could presumably check using some calculation with lie algebras but that doesn't sound fun or enlightening.

For $\mathbb{O}$ we could also do the following. The effect of $x\mapsto e^{(\theta/2)u}xe^{(\theta/2)u}$ is the plane rotation by $\theta$ with respect to the 2D plane $\mathrm{span}(1,u)$. We can conjugate two of these plane rotations to get an arbitrary plane rotation of $\mathbb{O}$, which generates all of them in $SO(8)$.

Alternatively, note that $x\mapsto -\overline{x}$ (where $\overline{x}$ is conjugation - it negates the imaginary part while preserve the real part) is a reflection across the hyperplane orthogonal to the real axis, i.e. across the subspace of imaginary elements. If we conjugate this reflection by a rotation from $1$ to $q$ (e.g. left multiplication by $q$, where $q$ is a unit quaternion), we get $x\mapsto q(-\overline{qx})=-q\overline{x}q^{-1}$ which is reflection across the hyperplane orthogonal to $q$. If we do this for both $p$ and $q$, we get that $x\mapsto p(qxq^{-1})p^{-1}$ is some plane rotation.

This gives another way to see that multiplication on the left and right by unit octonions generates $SO(8)$. For quaternions, in fact we have a $2$-to-$1$ group homomorphism $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to SO(4)$.

Far less obviously, since $u^2=-|u|^2$ for imaginary octonions, we get a left action of the Clifford algebra $\mathrm{Cliff}(7)$ on $\mathbb{O}\cong\mathbb{R}^8$. From the classification of Clifford algebras, we know $\mathrm{Cliff}(7)\cong M_8(\mathbb{R})\oplus M_8(\mathbb{R})$. Since the algebra representation $\mathrm{Cliff}(7)\to\mathrm{End}_{\mathbb{R}}(\mathbb{O})$ is nontrivial, by checking dimensions and using the fact the only ideals are the two summands we see the map is surjective. Therefore, every rotation of $\mathbb{O}$ can be attained simply by left multiplying by enough pure imaginary octonions. (The same works on the right side as well.)