A general strategy is to compute the expected number of games $t(xy)$ played until a team is declared the winner, starting from every possible partial score of $x$ games won by a player vs $y$ games won by the other one. Then the expected total number of games is $t(00)$.
Some simple remarks: (i) $t(xy)=t(yx)$ by symmetry; (ii) $t(x4)=0$ for every $0\leqslant x\leqslant3$; (iii) looking at the result of the first game played yields a relation between $t(xy)$ and $t((x+1)y)$ and $t(x(y+1))$, for each $xy$.
Starting from the highest possible partial scores and going backwards, one gets successively, using remarks (i), (ii) and (iii),
$$t(33)=1,\quad t(23)=1+\tfrac12t(33)=\tfrac32,\quad t(22)=1+t(23)=\tfrac52,
$$
$$t(13)=1+\tfrac12t(23)=\tfrac74,\quad t(03)=1+\tfrac12t(13)=\tfrac{15}8,
$$
$$
t(12)=1+\tfrac12t(13)+\tfrac12t(22)=\tfrac{25}8,\quad t(02)=1+\tfrac12t(03)+\tfrac12t(12)=\tfrac72,
$$
$$
t(11)=1+t(12)=\tfrac{33}8,\quad t(01)=1+\tfrac12t(11)+\tfrac12t(02)=\tfrac{77}{16},
$$
and, finally, $t(00)=1+t(01)=\frac{93}{16}$.
Edit: To check this result, note that
$$
t(00)=\frac{2{3\choose 0}2^3\cdot4+2{4\choose 1}2^2\cdot5+2{5\choose 2}2^1\cdot6+{6\choose 3}2^0\cdot7}{2{3\choose 0}2^3+2{4\choose 1}2^2+2{5\choose 2}2^1+{6\choose 3}2^0}.
$$
Let $a$ be the probability that the first player (ultimately) wins if the two players are tied in wins. Let $b$ be the probability that she wins if she is $1$ ahead. And let $c$ be the probability she wins if she is $1$ behind. We have the equations
$$\begin{align}a&=rb+(1-r)c,\\
b&=r+(1-r)a, \\
c&=ra.\end{align}$$
Solve the system of linear equations.
Best Answer
Let $a$ be the expected number of additional games that need to be played if the two players are tied in wins. In particular, $a$ is our required number, since the players start off tied.
Let $b$ be the expected number of additional games if Player $1$ is leading by $1$. And let $c$ be expected number of additional games if Player $1$ is trailing by $1$. We have the equations $$\begin{align}a&=1+rb+(1-r)c,\\ b&=1+(1-r)a, \\ c&=1+ra.\end{align}$$
Use the above system of linear equations to find $a$.
Remarks: $1.$ To justify the first equation, note that for sure we will be playing one more game. That's the $1$ in the equation. And we will not be finished. With probability $r$ our expected number of games beyond that will be $b$, and with probability $1-r$ our expected number of games beyond that will be $c$. That yields the first equation. It may be prettier and clearer to write the equation as $a=r(1+b)+(1-r)(1+c)$.
The justifications for the other two equations are similar. Again, one might like to write the second equation as $b=r(1)+(1-r)(a+1)$, and do a similar rewrite of the third equation.
$2.$ As pointed out by Marc van Leeuwen, the argument tacitly assumes that the expectations exist. To show that they do is not difficult. Whatever $r$ is, the probability of two opposite results in a row is $2r(1-r)$, which is $\le 1/2$. So the probability that a game length $2n$ or more is $\le (1/2)^n$, and therefore the expected length is finite.
$3.$ We used a strategy broadly similar to the one used in the related question about the probability that Player $1$ wins. This kind of strategy tends to work more widely for expectations than for probabilities, because of the linearity of expectation.