1. Let $S=(a+b)-\frac{a^2+b^2}{2}+\frac{a^3+b^3}{3}-\frac{a^4+b^4}{4}+\cdots$

2. Recognize the standard series: $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots, \qquad |x|<1$

   Since $0<a,b<1$, this expansion is valid for both $a$ and $b$.

3. Therefore, $a-\frac{a^2}{2}+\frac{a^3}{3}-\frac{a^4}{4}+\cdots = \ln(1+a)$ and $b-\frac{b^2}{2}+\frac{b^3}{3}-\frac{b^4}{4}+\cdots = \ln(1+b)$

   So, $S=\ln(1+a)+\ln(1+b)=\ln\big((1+a)(1+b)\big)$

4. Now use the given condition: $\tan^{-1} a + \tan^{-1} b = \frac{\pi}{4}$

   Taking tangent on both sides, $\tan\left(\tan^{-1} a + \tan^{-1} b\right)=\tan\frac{\pi}{4}=1$

   Using $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$ we get $\frac{a+b}{1-ab}=1$

   Hence, $a+b=1-ab$ $a+b+ab=1$

5. Then, $(1+a)(1+b)=1+a+b+ab=1+1=2$

6. Therefore, $S=\ln\big((1+a)(1+b)\big)=\ln 2$

7. Comparing with the options, this is: $\log_e 2$

Hence the correct option is A.