The condition $aH=Ha$ characterizes a special kind of subgroup, called a normal subgroup. They play a very important role.
Theorem. Let $G$ be a group, and let $H$ be a subgroup of $G$. The following are equivalent:
- Every left coset of $H$ is also a right coset of $H$.
- $aH = Ha$ for every $a\in G$.
- $aHa^{-1} = H$ for every $a\in G$.
- $aHa^{-1}\subseteq H$ for every $a\in G$.
- The equivalence relations $a\equiv_H b$ and $a {}_H\equiv b$ are the same equivalence relation.
- There exists a group $K$ and a homomorphism $f\colon G\to K$ such that $H=\mathrm{ker}(f)$.
(Remember that $a\equiv_H b$ if and only if $ab^{-1}\in H$; and $a{}_H\equiv b$ if and only if $a^{-1}b\in H$).
If $H$ is a normal subgroup, then your implication holds, exactly the same way as it holds for abelian groups.
Conversely, suppose that $H$ is not a normal subgroup. Then there exists an $a$ such that $aH\neq Ha$; but $a\in aH\cap Ha$. Thus, $aH\neq Ha$ but $aH\cap Ha\neq\emptyset$, so the implication does not hold. That is: your desired implication is equivalent to $H$ being a normal subgroup of $G$.
So we could add a seventh point to the theorem: "if $aH\cap Hb\neq\emptyset$, then $aH=Hb$."