I have the following equation:
$r_{*}=r+\frac{r_{+}^{2}}{\left(r_{+}-r_{-}\right)} \ln \left|r-r_{+}\right|-\frac{r_{-}^{2}}{\left(r_{+}-r_{-}\right)} \ln \left|r-r_{-}\right|$
and I need to solve for $r$, I am trying to do so in terms of known functions in hope to make the solution less numerically expensive.
For example, if $r_{-}$ is zero, I can solve for $r$ as:
$r = r_{+}\left (W\left(\frac{e^{r_{*}/r_{+} - 1}}{r_{+}}\right) + 1\right) $, where $W$ is the Lambert $W$ function.
In the case where $r_{-}$ is not zero I have not been able to do something similar. Is this even possible? maybe in terms of other functions?
I would very much appreciate any input, or ideas. Thank you for your time
$r_{-}=0$, 
For personal conveniency, I shall change notations : $a=r_+$, $b=r_-$ and $y=r_*$. To get rid of the absolute values, I shall use $\log(|t|)=\frac 12 \log(t^2)$ and do not make any hypothesis about the signs of $a$ and $b$ but assume $a>b$.
So, the equation write $$y=r+\frac{a^2 }{2 (a-b)}\log \left((r-a)^2\right)-\frac{b^2 }{2(a-b)}\log \left((r-b)^2\right)$$ Since $a>b$, I would generate an estimate ignoring the last term. As you observed, if $b=0$, we have a solution which is now $$r_0=a+\frac {a^2}{a-b}\,W(\pm \sqrt t)\quad \text{with} \quad t=\frac{(a-b)^2}{a^4}\exp\Bigg[ \frac{2 (a-b) (y-a)}{a^2}\Bigg]$$
For a simple test, I used $a=2\pi$, $b=e$ and $y=50$. This gives $r_0 \sim 20.558733$.
Now, Newton iterates $$\left( \begin{array}{cc} n & r_n \\ 0 & 20.558733 \\ 1 & 24.157560 \\ 2 & 24.331464 \\ 3 & 24.331763 \end{array} \right)$$