Using algebraic topology, one can in a relatively straightforward matter the following:

**Proposition** For even $n>0$, there does not exist a map $\mu\colon S^n\times S^n\to S^n$ such that $$\mu\circ i_1=\mathrm{id}_{S^n}=\mu\circ i_2,$$ where $i_1,i_2\colon S^n\to S^n\times S^n$ are (continuous) inclusions of the form $S^n\to S^n\times\{x_1\}$ and $S^n\to\{x_2\}\times S^n$.

Whether algebraic topology is less elementary than using von Neumann's theorem to pass to Lie groups, giving a parallel vector field, and then going through the hairy ball theorem (or its higher dimension analogs) is... debatable. Nevertheless, the algebraic topology result is more general since it shows the impossibility not only of topological group structure (by taking $x_1=x_2=e$), but of the structure of a topological magma (i.e. an H-space).

*Proof.* The Künneth formula tells us that the (unital) cohomology algebra $H^*(S^n\times S^n)$ is isomorphic to the tensor product $H^*(S^n)\otimes H^*(S^n)$, with the isomorphism given by $$H^*(S^n)\otimes H^*(S^n)\ni a\otimes b\mapsto \pi_1^*(a)\cup\pi_2^*(b)\in H^*(S^n)$$ where $\pi_1,\pi_2\colon S^n\times S^n\to S^n$ are the canonical projection maps. In particular, if $a$ is a generator for $H^*(S^n)$ (as a unital algebra), then $\pi_1^*(a)$,$\pi_2^*(a)$ are (independent) generators of $H^*(S^n\times S^n)$ as a unital algebra, and of $H^n(S^n\times S^n)$ as a $2$-dimensional vector space.

Now, the continuous inclusions $i_1,i_2\colon S^n\to S^n\times S^n$ are maps such that $$\pi_i\circ i_j=\begin{cases}\mathrm{id}_{S^n}&i=j\\\text{constant}&i\neq j\end{cases}.$$ Therefore the induced homomorphisms $i_1^*,i_2^*\colon H^*(S^n\times S^n)\to H^*(S^n)$ and $\pi_1^*,\pi_2^*\colon H^n(S^n)\to H^*(S^n\times S^n)$ of cohomology algebras are such that $$i_j^*\circ\pi_i^*=\begin{cases}\mathrm{id}_{H^*(S^n)}&i=j\\0&i\neq j\end{cases}.$$ In particular, if $$\sigma=x\pi_1^*(a)+y\pi_2^*(a)\in H^n(S^n\times S^n),$$ then $i_1^*(\sigma)=xa$ and $i_2^*(\sigma)=ya$.

Consequently, if $\mu\colon S^n\times S^n\to S^n$ is such that $\mu\circ i_j=\mathrm{id}_{S^n},$ then the induced map $\mu^*\colon H^*(S^n)\to H^*(S^n\times S^n)$ on cohomology algebras must satisfy $$i_j^*\circ\mu^*=\mathrm{id}_{H^*(S^n)}\colon H^*(S^n)\to H^*(S^n).$$ We have that $i_j^*(\mu^*(a))=a$ for $\mu^*(a)\in H^n(S^n)$, so from the preceding observation $\mu^*(a)=\pi_1^*(a)+\pi_2^*(a)$.

On the other hand, $a\cup a\in H^{2n}(S^n)=0$, so we have
\begin{align*}0=\mu^*(0) &=\mu^*(a\cup a) \\
&=\mu^*(a)\cup\mu^*(a) \\
&=(\pi_1^*(a)+\pi_2^*(a))^2 \\
&=\pi_1^*(a\cup a)+\pi_2^*(a\cup a)+\pi_1^*(a)\cup\pi_2^*(a)+\pi_2^*(a)\cup\pi_1^*(a) \\
&=\pi_1^*(a)\cup\pi_2^*(a)+(-1)^n\pi_1^*(a)\cup\pi_2^*(a).
\end{align*} Since $$\pi_1^*(a)\cup \pi_2^*(a)\cong a\otimes a\neq0\in H^*(S^n\times S^n)\cong H^*(S^n)\otimes H^*(S^n),$$ this equality holds if and only if $n$ is odd.

**Remark.** I think (but I have not tried this) that if we assume further that $\mu\colon S^n\times S^n\to S^n$ is an associative multiplication, then applying an analogous argument to $$\mu^{(k)}\colon S^n\times S^n\times\dots\times S^n\to S^n$$ might give extra conditions on the dimension. One could maybe show in this way that $H^*(S^n)$ can have an induced structure as a Hopf algebra (i.e. associative multiplication, and an inversion map) only if the dimension is $n=2^m-1$. Or maybe not, I don't know.