regular expression of star-height 1
The language has star-height 2. Note that the language includes
$R_0=(ba^+ba^+)^*$, but is disjoint from $R_1=(ba^+ba^+)^*ba^+$. Therefore, it suffices to show:

Claim: If $L$ has star-height 1 and $R_0\subseteq L$, then $L\cap R_1\ne\emptyset$.

Suppose $L$ has star-height 1. Then we can write $L=\bigcup_{i=1}^m F_{i,0}^* w_{i,1} F_{i,1}^* \cdots w_{i,n} F_{i,n}^*$ for words $w_{i,j}$ and finite sets $F_{i,j}$. 

We have $(ba^kba^k)^k\in R_0\subseteq L$ for every $k$. For $k$ large enough,
there is some $F_{i,j}^*$ that contributes at least three $b$'s. Let us
choose $k$ so large that also $k&gt;|w|$ for any $w\in \bigcup_{i,j} F_{i,j}$.

Let $u$ be the factor with at least three $b$'s contributed by $F_{i,j}^*$. Then $u$ decomposes into factors $u_1,\ldots,u_m$ from $F_{i,j}$. Now we can
clearly pick one or two consecutive $u_\ell$'s that together form a word $f\in a^+ba^+$. We can therefore repeat $f$ and obtain a word in $R_1$.

Update: Tried to be clearer.