Is each preseparable topological group narrow?

A topological group $$G$$ is defined to be

$$\bullet$$ precompact if for any neighborhood $$U\subseteq G$$ of the unit there exists a finite subset $$F\subseteq G$$ such that $$G=UF$$;

$$\bullet$$ narrow if for any neighborhood $$U\subseteq G$$ of the unit there exists a countable subset $$S\subseteq G$$ such that $$G=US$$;

$$\bullet$$ separable if there exists a countable subset $$S\subseteq G$$ such that for any neighborhood $$U\subseteq G$$ of the unit we have $$G=SU$$;

$$\bullet$$ preseparable if there exists a countable subset $$S\subseteq G$$ such that for any neighborhood $$U\subseteq G$$ of the unit there exists a finite subset $$F\subseteq G$$ such that $$G=SUF$$.

Let us observe the following facts concerning those concepts:

1. A topological group is preseparable if it is precompact or separable.

2. More generally, a topological group $$G$$ is preseparable if $$G$$ contains a separable closed normal subgroup $$H$$ whose quotient group $$G/H$$ is precompact. In the latter case the group $$G$$ is also narrow.

3. Each preseparable abelian topological group is narrow.

4. For any cardinal $$\kappa>\mathfrak c$$, the Tychonoff power $$\mathbb R^\kappa$$ is an example of a narrow abelian topological group which is not preseparable.

Problem. Is each preseparable topological group narrow?

Theorem. If a group $$G$$ is written as $$G=\bigcup_{i=1}^nU_iA$$ for some sets $$U_1,\dots,U_n,A\subset G$$, then $$G=U_i^{-1}U_iB$$ for some $$i\in\{1,\dots,n\}$$ and some set $$B\subseteq G$$ of cardinality $$|B|\le f(n,|A|)$$, where the function $$f(n,\kappa)$$ is defined by the recursive formula: $$f(1,\kappa)=\kappa$$ and $$f(n,\kappa)=f(n-1,\kappa+\kappa^2)$$. In particular, $$f(n,\kappa)=\kappa$$ for any infinite cardinal $$\kappa$$ and any $$n\in\mathbb N$$.
Proof. The proof is by induction on $$n$$. For $$n=1$$ it is trivial. Assume that the theorem is proved for all $$k. Write $$G$$ as $$G=\bigcup_{i=1}^nU_iA$$ for some sets $$U_1,\dots,U_n,A\subset G$$. If $$U_n^{-1}U_nA=G$$, then we are done. If $$U_n^{-1}U_nA\ne G$$, then we can choose a point $$x\in G\setminus U_n^{-1}U_nA$$ and conclude that $$U_nx\cap U_nA=\emptyset$$ and hence $$U_nx\subset \bigcup_{i=1}^{n-1}U_iA$$. Then $$U_nA\subset \bigcup_{i=1}^{n-1}U_iAx^{-1}A$$ and $$G=\bigcup_{i=1}^{n-1}U_i(A\cup Ax^{-1}A)$$. By the induction hypothesis, there exists $$i\in\{1,\dots,n-1\}$$ and a set $$B\subset G$$ of cardinality $$|B|\le f(n-1,|A\cup Ax^{-1}A|)\le f(n-1,|A|+|A|^2)=f(n,|A|)$$ such that $$G=U_i^{-1}U_iB$$.