Certainly the strategy for each player must be of the form to reroll below a certain threshold. To find the threshold, we can start from Ross’ approximate solution and iteratively adjust the strategies until they're in equilibrium.
So assume that $B$ rerolls below $6$ and $A$ rerolls below $9$. Then every number of $B$ below $6$ has probability $\frac12\cdot\frac1{10}=\frac1{20}$, and every number $6$ or higher has probability $\frac1{10}+\frac12\cdot\frac1{10}=\frac3{20}$. Every number of $A$ below $9$ has probability $\frac8{20}\cdot\frac1{20}=\frac1{50}$ and every number $9$ or higher has probability $\frac1{20}+\frac8{20}\cdot\frac1{20}=\frac7{100}$.
If $B$ were to adjust her strategy by keeping $5$, she would get a winning chance of $5\cdot\frac1{50}=\frac1{10}$ instead of the current winning chance of $\frac1{10}\sum_{k=1}^{10}\frac k{50}+3\cdot\frac1{10}\left(\frac7{100}-\frac1{50}\right)$, where the second term corrects for the $9$ and $10$ that are incorrectly included in the sum in the first term. This is $\frac{10(10+1)}{2\cdot10\cdot50}+\frac3{200}=\frac{11}{100}+\frac3{200}=\frac{25}{200}=\frac18$. Thus, there is no gain in this strategy change.
On the other hand, if she were to adjust her strategy by rerolling $6$, she would give up the current winning probability of $6\cdot\frac1{50}=\frac3{25}$ to get the above winning probability upon rerolling, $\frac18$. Since this is slightly larger than $\frac3{25}$, she should switch strategies and reroll $6$. (The reason this didn't show up in Ross’ argument is that given $A$'s strategy, $9$ and $10$ are far more likely than the other numbers, so it makes sense for $B$ to try harder to beat them even though it lowers $B$'s average roll.)
On the other hand, keeping $7$ yields a winning probability of $7\cdot\frac1{50}=\frac7{50}\gt\frac18$, so $B$ shouldn't reroll $7$. Thus $B$’s best response to $A$’s current strategy is to reroll $6$ or lower. The probability for $B$ to end up with a number below $7$ will then be $\frac6{10}\cdot\frac1{10}=\frac3{50}$, and the probability for a number $7$ or higher will be $\frac1{10}+\frac6{10}\cdot\frac1{10}=\frac4{25}$.
Now we should check whether $A$’s strategy is the best reponse to $B$’s new strategy. If $A$ were to adjust his strategy by keeping $8$, he’d get a winning chance of $1-3\cdot\frac4{25}=\frac{13}{25}$ instead of the current winning chance of $\frac12+\frac1{20}\sum_{k=1}^{10}(k-1)\cdot\frac3{50}+6\cdot\frac1{20}\left(\frac4{25}-\frac3{50}\right)=\frac12+\frac{3\cdot9\cdot10}{2\cdot20\cdot50}+\frac3{100}=\frac12+\frac{27}{200}+\frac3{100}=\frac{133}{200}$. Since $\frac{13}{25}=\frac{104}{200}$ is lower, there is no gain in this strategy change.
On the other hand, keeping $9$ yields a winning probability of $1-2\cdot\frac4{25}=\frac{17}{25}=\frac{136}{200}\gt\frac{133}{200}$, so rerolling $9$ is no gain, either. Thus, $A$’s current strategy of rerolling below $9$ is still the best response to $B$’s new strategy. The overall winning probability of $A$ for this equilibrium strategy pair is
$$
\frac12+\frac1{20}\left(1-\frac4{25}+1-\frac8{25}+8\left(\frac12+\frac1{20}\left(1-\frac4{25}+1-\frac8{25}+\sum_{k=1}^8(k-1)\cdot\frac3{50}\right)\right)\right)
\\
=
\frac12+\frac1{20}\left(\frac{38}{25}+4+\frac25\left(\frac{38}{25}+\frac{7\cdot8\cdot3}{2\cdot50}\right)\right)=\frac{21}{25}=0.84\;,
$$
a slight drop from Ross’ approximation due to the slight improvement in $B$’s strategy.
Best Answer
As I outlined in a comment, we denote by $t_k, 1 \leq k \leq 6$, the expected number of further rolls required to get six distinct numbers when there are currently $k$ distinct numbers. So, for example, $t_6 = 0$, because there are already six distinct numbers.
If there are currently $k = 5$ distinct numbers, we pick up the duplicate and reroll it. There are two possibilities:
This allows us to write the recurrence
$$ t_5 = 1 + \frac56 t_5 $$
which can be solved to yield $t_5 = 6$. If there are $k = 4$ distinct numbers, we pick up the two duplicates and reroll them. There are four possibilities:
The second and third cases both yield five distinct numbers, so we can write the recurrence
$$ t_4 = 1 + \frac49 t_4 + \left(\frac49+\frac{1}{18}\right) t_5 $$
Plugging in $t_5 = 6$ reduces this to
$$ t_4 = 4 + \frac49 t_4 $$
which yields $t_4 = \frac{36}{5}$. In general, we may write
$$ t_k = 1 + \sum_{j=k}^5 p_{kj} t_j $$
where $p_{kj}$ is the probability of going from $k$ distinct numbers to $j \geq k$ distinct numbers in a single roll. There's probably an explicit summation form for this, but I'm afraid I'm too lazy to think of it at the present time. At any rate, we can continue along in the same vein to write
\begin{align} t_3 & = 1 + \frac18 t_3 + \frac{37}{72} t_4 + \frac13 t_5 \\ & = 1 + \frac18 t_3 + \frac{37}{10} + 2 \\ & = \frac{67}{10} + \frac18 t_3 \end{align}
yielding $t_3 = \frac{268}{35}$, then
\begin{align} t_2 & = 1 + \frac{1}{81} t_2 + \frac{65}{324} t_3 + \frac{55}{108} t_4 + \frac{7}{27} t_5 \\ & = 1 + \frac{1}{81} t_2 + \frac{871}{567} + \frac{11}{3} + \frac{14}{9} \\ & = \frac{4399}{567} + \frac{1}{81} t_2 \end{align}
yielding $t_2 = \frac{4399}{560}$, then
\begin{align} t_1 & = 1 + \frac{1}{7776} t_1 + \frac{155}{7776} t_2 + \frac{25}{108} t_3 + \frac{325}{648} t_4 + \frac{25}{108} t_5 \\ & = 1 + \frac{1}{7776} t_1 + \frac{136369}{870912} + \frac{335}{189} + \frac{65}{18} + \frac{25}{18} \\ & = \frac{986503}{124416} + \frac{1}{7776} t_1 \end{align}
yielding $t_1 = \frac{986503}{124400}$. (Thanks to @user in the comments for noticing an error in my original computation!) Finally, we observe that if $k = 1$ (that is, if you only have one distinct number), you're essentially right where you started, so the overall expected number of rolls until you get six distinct numbers is
$$ t = t_1 = \frac{986503}{124400} \approx 7.93009 $$
There may be a simpler and cleverer way to this answer.