$\def\RR{\mathbb{R}}$This is an answer to flesh out what I wrote in comments. Let $V$ be a real vector space with positive definite inner product. For a unit vector $\vec{u} \in V$, let $s_{\vec{u}}$ be the reflection over $\vec{v}$. Here is something we might hope for:
Vague hope: Can we build a ring $R$ whose additive structure comes from the vector space $V$ and whose multiplicative structure comes from multiplying the reflections $s_{\vec{u}}$ in the orthogonal group $O(V)$?
Note that, in particular, if we want multiplication to come from $O(V)$, we should have $\vec{u}^2 = 1$.
More precisely,
Precise idea: Can we find an $\RR$-algebra $R$ which contains $V$ as a vector subspace, such that $\vec{u}^2=1$ for every unit vector $\vec{u} \in V$?
Well, if we ask for a linear map from $V$ to $R$ rather than insisting on a subspace, we can clearly do this. We simply define $R$ by generators and relations. This is why I don't understand the OP's objection to generators and relations: They are the correct tool whenever you want to build an algebraic object that has certain elements in it with certain properties.
Definition: Let $R$ be the $\RR$-algebra generated by $V$, subject to the relations $\vec{u}^2 = 1$ for every unit vector $\vec{u}$ in $V$.
The ring $R$ is the Clifford algebra, and it clearly has an $\RR$-linear map $V \to R$. It isn't obvious that this map is injective, but that is a lemma which you can find in any textbook on Clifford algebras, so I'll omit it. (The more precise statement is that, if $e_1$, $e_2$, ..., $e_n$ is any basis for $V$ then the $2^n$ products $e_{i_1} e_{i_2} \cdots e_{i_k}$ are a basis of $R$.)
Let's connect this to the standard definition. For any vector $\vec{v} \in V$, we can write $\vec{v} = c \vec{u}$ where $\vec{u}$ is a unit vector and $c \in \RR_{\geq 0}$. Since we asked $R$ to be an $\RR$-algebra, we have $\vec{v}^2 = (c \vec{u})^2 = c^2 = |\vec{v}|^2$. For any $\vec{v}$ and $\vec{w}$, we then have
$$\vec{v}^2 = \vec{v} \cdot \vec{v},\ \vec{w}^2 = \vec{w} \cdot \vec{w},\ \mbox{and}\ (\vec{v}+\vec{w})^2 = (\vec{v}+\vec{w}) \cdot (\vec{v}+\vec{w}).$$
Subtracting the first and second equation from the third,
$$\vec{v} \vec{w} + \vec{w} \vec{v} = 2 \vec{v} \cdot \vec{w}. \qquad (\ast)$$
Here the left hand side is multiplication in the ring $R$ and the right hand side is dot product. That's the more usual definition of the Clifford algebra. (Some people choose the opposite sign convention, $\vec{v} \vec{w} + \vec{w} \vec{v} = - 2 \vec{v} \cdot \vec{w}$. I don't have a firm opinion, but I started writing the answer this way, so I'll stick to it.)
Then a miracle happens! For any unit vector $\vec{u}$ and any vector $\vec{v}$, write $\vec{v} = \vec{v}^{\parallel} + \vec{v}^{\perp}$, where $\vec{v}^{\parallel}$ is parallel to $\vec{u}$ and $\vec{v}^{\perp}$ is orthogonal. We have $\vec{u} \vec{v}^{\parallel} = \vec{u} \cdot \vec{v} = \vec{v}^{\parallel} \vec{u}$, because $\vec{v}^{\parallel} = (\vec{u} \cdot \vec{v}) \vec{u}$ and $\vec{u}^2 = 1$. Equation $(\ast)$ also gives $\vec{u} \vec{v}^{\perp} = - \vec{v}^{\perp} \vec{u}$. So we have
$$\vec{u} \vec{v} = \vec{u} \left( \vec{v}^{\parallel} + \vec{v}^{\perp} \right) = \left( \vec{v}^{\parallel} - \vec{v}^{\perp} \right) \vec{u} = - s_{\vec{u}}(\vec{v}) \vec{u}. \qquad (\dagger)$$
Let $\vec{u}_1$, $\vec{u}_2$, ..., $\vec{u}_k$ be any sequence of reflections and let $g$ be the product $s_{\vec{u}_1} s_{\vec{u}_2} \cdots s_{\vec{u}_k}$ in the orthogonal group $O(V)$. Then using $(\dagger)$ repeatedly gives
$$\vec{u}_1 \vec{u}_2 \cdots \vec{u}_k \vec{v} = (-1)^k g(\vec{v}) \vec{u}_1 \vec{u}_2 \cdots \vec{u}_k. \qquad (!)$$
Let $\mathrm{Pin}(V)$ be the subgroup of the unit group of $R$ generated by the unit vectors $\vec{u}$. It is easy to see that there is a group map $\texttt{sign} : \mathrm{Pin}(V) \to \{ \pm 1 \}$ with $\texttt{sign}(\vec{u}) = -1$ for every unit vector. Then equation $(!)$ tells us that there is a map $\pi: \mathrm{Pin}(V) \to O(V)$ with $\pi(x)(\vec{v}) = \texttt{sign}(x) x v x^{-1}$.
In this sense, we have fulfilled our hope: $V$ is a vector subspace of $R$, and the fact that $\pi$ is a map of groups tells us that any multiplicative identities that hold between the $\vec{u}$ in $R^{\times}$ also hold between the $s_{\vec{u}}$ in $O(V)$.
The ring $R$ decomposes as $R_+ \oplus R_-$ where $R_+$ is spanned by products of an even number of vectors and $R_-$ uses an odd number. The subgroup $\mathrm{Spin}(V) = \mathrm{Pin}(V) \cap R_+$ maps to $SO(V)$, so this is where to look for rotations.
So far, $V$ can have any dimension. Now we specialize to $\RR^3$, with orthogonal basis $e_1$, $e_2$, $e_3$. It is easy to check that the elements $e_1 e_2$, $e_2 e_3$ and $e_1 e_3$ multiply by the same relations as $i$, $j$ and $k$ in the quaternions and, in fact, $R_+ \cong \mathbb{H}$.
The special magic thing that happens in the quaterions is that $\mathrm{Spin}(\RR^3)$ is the unit sphere in $\mathbb{H}$. In general, $\mathrm{Spin}(\RR^n)$ is an $\binom{n}{2}$-dimensional submanifold of a $2^{n-1}-1$ dimensional unit sphere in $R_+$. It is only when $n=1$, $2$ or $3$ that $\binom{n}{2} = 2^{n-1} - 1$.
