Let $G$ be a locally compact group and $(a_n),(b_n)$ be two sequences.
Suppose $a_n b_n \to e$ as $n \to \infty$，do we necessarily have $b_n a_n \to e$?
If this is false in general, does it hold true for $G=SL(d,\mathbb R)$?
The group I am mostly interested in is $SL(d,\mathbb R)$. I have been trying to construct a counterexample here, but not successful.
On
Let $G$ an arbitrary metrizable topological group.
Then the answer of your question is yes if and only if $G$ has a basis of conjugacy-invariant neighborhoods.
Indeed, suppose the latter holds. Let $a_nb_n\to 1$. Let $V$ be a neighborhood of $1$. There exists a conjugacy-invariant neighborhood $V$ of $1$ contained in $V$. Then for $n$ large enough, say $n\ge N$, $a_nb_n\in W$. By conjugacy-invariance, $b_na_n\in W$ for all $n\ge N$. So $b_na_n\to 1$.
Conversely, fix a metric on $G$ defining the topology, and let $W_n$ be the smallest conjugacy-invariant subset containing the $1/n$-ball around $1$. Suppose, by contraposition, that $W_n$ is not a basis of neighborhoods of $1$. Then there exists $r>0$ such that for every $n$ there exists some element $h_n\in W_n$ at distance $\ge r$ from $1$. Since $h_n\in W_n$, there exists $g_n$ such that $g_nh_ng_n^{-1}$ has distance $\le 1/n$ to $1$. Then, for $a_n=g_nh_n$ and $b_n=g_n^{-1}$ we have $a_nb_n=g_nh_ng_n^{-1}\to 1$ while $b_na_n=h_n$ does not tend to $1$.
Indeed most non-abelian Lie groups do not satisfy this property and this yields many counterexamples.
Consider the following matrices in the case where $d=2$.
$A_n=\begin{pmatrix} 1/n & 0 \\ 1/n^2 & n \end{pmatrix}$ and $B_n=\begin{pmatrix} n & 1 \\ 0 & 1/n \end{pmatrix}$.
Both of them are invertible with unit determinant, hence in $SL(2, \mathbb{R})$.
You have $A_n B_n = \begin{pmatrix} 1 & 1/n \\ 1/n & 1+1/n^2 \end{pmatrix} \to I_2$ while $B_nA_n = \begin{pmatrix} 1+1/n^2 & n \\ 1/n^3 & 1 \end{pmatrix}$.
You see that $B_nA_n$ does not converge to $I_2$.