Hypothesis: Let $G$ be a topological group. Let $\mu$ denote the group multiplication mapping on $G \times G$ to $G$ that is stipulated to be continuous. Let $\pi_1(G,e) = \pi(G)$ denote the fundamental group of $G$.
Question: I've read elsewhere that there is a product $*$ induced by
$$ * : \pi(G) \times \pi(G) \cong \pi(G \times G) \overset{\pi(\mu)}{\rightarrow} \pi(G) $$
I am confused by what exactly is meant here. This product is distinct from the usual group product $\circ$ on $\pi(G)$. If anything, what is meant by $\pi(\mu)$?
The map $\mu\colon G\times G\to G$ is a continuous map. The map $\pi(\mu)\colon \pi(G)\times \pi(G)\to \pi(G)$ is the induced group homomorphism on fundamental groups.
As an example, if $G=S^1$ with complex multiplication as the group operation $\mu\colon S^1\times S^1\to S^1$, then $\pi(S^1)=\mathbb{Z}$ and $\pi(S^1\times S^1)=\mathbb{Z}\oplus\mathbb{Z}$. The induced map $\pi(\mu)\colon \mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}$ maps $(u,v)$ to $u+v$.