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
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.