Let $G$ be an algebraic group. The product group $G \times G$ (taken as a product of varieties) contains the product topology, and is a product with respect to the canonical projections $G \times G \rightarrow G$.
Fix $g \in G$, and let $f: G \rightarrow G$ be given by $f(x) = xg$. Why is this a homeomorphism of $G$? I don't even see why this should be continuous. It would be continuous by the composition $$G \rightarrow G \times \{g\} \rightarrow G \times G \xrightarrow{\mu} G$$ where $\mu$ is the multiplication map (by definition continuous), the middle arrow is the inclusion map of $G \times \{g\}$ (continuous if we take $G \times \{g\}$ in the subspace topology), and the left arrow as the map $x \mapsto (x,g)$. However, I cannot guarantee that this left arrow is continuous.
The left arrow is a distraction. This map is the composite
$$G \xrightarrow{\text{id}_G \times g} G \times G \xrightarrow{m} G$$
where $m$ is the multiplication, $g$ is the map $1 \to G$ corresponding to $g \in G$, and for two maps $f : A \to B, g : A \to C$, by $f \times g$ I mean the map $A \to B \times C$ whose components (the composites $\pi_B \circ f, \pi_C \circ f$ given by projecting to $B$ or $C$) are $f$ and $g$ respectively.
If $G$ is a group object in a category with finite products, then this composite is a well-defined morphism in that category for purely formal reasons. Hence $x \mapsto gx$ is not only continuous, it is in fact algebraic, and it has inverse $x \mapsto g^{-1} x$ as one might expect.
The only remaining question is why an algebraic map from a variety to itself is continuous on underlying topological spaces, and that has nothing to do with groups or with the difference between the Zariski topology and the product topology.