Let $A \subseteq B(H)$ be a $*$-subalgebra of the bounded operators on the Hilbert space $H$. Let $B$ be the SOT-closure of $A$. Is $B$ again abelian?
Let $u,v \in B$. We can find nets $(u_\lambda), (v_\lambda)$ in $A$ with $u_\lambda \to u$ and $v_\lambda \to v$ in the strong topology. The obvious guess would be that $u_\lambda v_\lambda \to uv$ but in general this is not true since the multiplication map is not strongly continuous, so we will need to make another approach.
On
In case $A$ is unital there is another approach: von Neumann's double commutant Theorem says that the strong closure of $A$ coincides with $A''$, where the commutant $B'$ of a subset $B\subseteq B(H)$ is defined by $$ B'=\{T\in B(H): TS=ST, \text{ for all } S\in B\}. $$
The fact that $A$ is abelian translates into $A\subseteq A'$ and since the operation of taking the commutant is clearly inclusion-reversing, we have that $A''\subseteq A'$. Applying this once more we get $A''\subseteq A'''$ which precisely says that $A''$ (a.k.a. the strong closure of $A$), is commutative.
But you can do it in two steps. For any $a\in A$, $ua=\lim u_\lambda a=\lim au_\lambda=a\lim u_\lambda=au$. Now $uv=\lim uv_\lambda=\lim v_\lambda = vu$.