How if we have a neighborhood $U$ of $1$, with $\overline{U^{-1}}$ compact?
It seems the following.
The answer is positive.
Let $\mathcal B$ be a base at the unit $1$ of the group $(G,\mathcal T)$. Put $B=\bigcap\{\overline{V^{-1}}:V\in\mathcal B\}$. Then $B$ is a compact closed subset of $G$. Since the multiplication of the group $G$ is continuous at the unit, for each neighborhood $V\in\mathcal B$ there exists a neighborhood $W\in\mathcal B$ such that $WW\subset V$. Then $W^{-1}W^{-1}\subset V^{-1}$. Since $\overline{V^{-1}}\subset V^{-1}V^{-1}$ for each neighborhood $V\in\mathcal B$, we see that $B=\bigcap\{V^{-1}:V\in\mathcal B\}$ is a compact semigroup. Then there exists a minimal closed right ideal $H\subset B$. Choose an arbitrary element $x\in H$. Then $x^2H=H\ni x$ hence $x^{-1}\in H$ and $1\in H$. Thus $H=B$ and $B$ is a subgroup of the group $G$. (In fact, since $B$ is compact, it is a topological group by my old result (see Proposition below).
Now suppose that $a,b\in G$ and each neighborhood of $a$ contains $b$. Let $V\in\mathcal B$ be an arbitrary neighborhood of the unit. Then $aV\ni b$, so $a\in bV^{-1}$. That is $a\in bB$ and $b\in aB^{-1}$. Since $B$ is a group, $B^{-1}=B$ so $b\in aB\subset aV^{-1}$. Then $bV\ni a$, that is each neighborhood of $b$ contains $a$.
clearly every closed neighborhood of $b$ contains $a$. So I wonder if any open neighborhood of $1$ contains a closed neighborhood of $1$.
This is another question, which has a negative answer. Consider the following counterexample. Define on the group $\Bbb Z$ of integers a topology $\mathcal T$ as follows. Let $p>1$ be a natural number. For each positive integer $n$ put $U_n=p^n(\Bbb N\cap\{0\})$. Put a family $\{m+U_n: m\in\Bbb Z, m\in\Bbb N\}$ as a base of the topology $\mathcal T$. It is easy to check that $(\Bbb Z,\mathcal T)$ is a Hausdorff paratopological group, but $\overline{U_n}=p^n\Bbb Z\not \subset U_1$ for each $n$.
Proposition. A compact paratopological group is a topological group.
Proof. Let $G$ be a compact paratopological group. At first we consider the case when $G$ is $T_1$. Let $C$ be a closed subset of the group $G$. Then $C^{-1}=\{x\in G:xC\ni e\}$. Let $x\not\in C^{-1}$. Then for every point $y\in C$ there exist open neighborhoods $Ox(y)\ni x$, $Oy\ni y$ such that $Ox(y)Oy\not\ni e$. Choose a finite subcover $\{Oy:y\in Y\}$ of the set $C$ and put $Ox=\bigcap\{Ox(y):y\in Y\}$. Then $OxC\not\ni e$ and therefore the set $C^{-1}$ is closed. Therefore the map $i:G\to G$, $i:x\mapsto x^{-1}$ is continuous.
Now consider the general case. Let $\mathcal B$ be a base at the unit $e$ of the group $G$. Put $B=\bigcap\mathcal B$. Since the multiplication of the group $G$ is continuous at the unit, for each neighborhood $U\in\mathcal B$ there exists a neighborhood $V\in\mathcal B$ such that $VV\subset U$. This implies that $B$ is a semigroup. Put $B'=\overline{\{e\}}=\{x\in G:\forall U\in\mathcal B$ $xU\ni e\}=\{x\in G:xB\ni e\}= B^{-1}$. Hence $B'$ is a closed semigroup of the group $G$. Since $B'$ is a compact semigroup then there exists a minimal closed right ideal $H\subset B'$. Choose an arbitrary element $x\in H$. Then $x^2H=H\ni x$ hence $x^{-1}\in H$ and $e\in H$. Thus $H=B'$ and $xB'=B'$ for every element $x\in B'$. Therefore $B'$ is a group and $B'=B$. Since $B=\bigcap\mathcal B$ then $g^{-1}Bg\subset B$ for every $g\in G$. Hence $B$ is a normal subgroup of the group $G$. The group $B$ is topological, because $B$ is endowed with the trivial topology. Since $G/B$ is $T_1$ compact topological group then Lemma implies that $G$ is a topological group. $\square$
Lemma. Let $G$ be a paratopological group and $H$ be a normal subgroup of the group $G$. If $H$ and $G/H$ are topological groups then $G$ is a topological group.
Proof. Let $U$ be an arbitrary neighborhood of the unit. There exist neighborhoods $V,W$ of the unit such that $V\subset U$, $(V^{-1})^2\cap H\subset U$ and $W\subset V$, $W^{-1}\subset VH$. If $x\in W^{-1}$ then there exist elements $v\in V,h\in H$ such that $x=vh$. Then $h=v^{-1}x\in V^{-1}W^{-1}\cap H\subset U$. Therefore $x\in VU\subset U^2$. Hence $G$ is a topological group.$\square$