I actually am not familiar with the method you're using, so I'm using a different method.
The function is actually concave on the (natural) domain $\Bbb{R} \times (0, \infty)$. Note that the function $(x_1, x_2) \mapsto \ln(x_2)$ is concave, because the function $\ln$ is concave (check its second derivative). The function $(x_1, x_2) \mapsto x_1$ is an affine function, and hence is concave (and convex). Summing two concave functions produces a concave function, and every concave function is quasiconcave.
The only question remaining is strictness. Neither of the above functions are strictly (quasi)concave, so we need a separate argument. Strict quasiconcavity means that, for all $(x_1, x_2), (y_1, y_2)$ in the domain, and any $\lambda \in (0, 1)$, we have
$$f(\lambda x_1 + (1 - \lambda)y_1, \lambda x_2 + (1 - \lambda)y_2) > \min \{f(x_1, x_2), f(y_1, y_2)\}.$$
Quasiconvexity means the above with $\ge $ substituted for $>$. So, let's suppose that we have equality. Without loss of generality, assume $f(x_1, x_2) \le f(y_1, y_2)$. Then,
$$\lambda x_1 + (1 - \lambda)y_1 + \ln(\lambda x_2 + (1 - \lambda)y_2) = x_1 + \ln(x_2).$$
By the strict concavity of $\ln$ (again, examine the second derivative), we have
\begin{align*}
&x_1 + \ln(x_2) > \lambda x_1 + (1 - \lambda)y_1 + \lambda\ln(x_2) + (1 - \lambda)\ln(y_2) \\
\iff \, &(1 - \lambda)x_1 + (1 - \lambda)\ln(x_2) > (1 - \lambda)y_1 + (1 - \lambda)\ln(y_2) \\
\iff \, &x_1 + \ln(x_2) > y_1 + \ln(y_2) \iff f(x_1, x_2) > f(y_1, y_2),
\end{align*}
which contradicts $f(x_1, x_2) \le f(y_1, y_2)$. Therefore, $f$ is quasiconcave.
Best Answer
AmerYR calculated $$f''(x)=\frac{a^2x^{a-1}\left(1+a+x^a(a-1)\right)}{(1-x^a)^3} - \frac{2}{(1-x)^3} \, .$$
As you noticed $f''(x)=0$ at $a=1$. So it suffices to show that $\partial_a f''(x) \leq 0$. In the comments I gave the derivative with respect to $a$ by $${\frac { \left( \left( \left( {a}^{2}-a \right) \ln \left( x \right) -3\,a+2 \right) {x}^{3\,a-1}+ \left( 4\,{a}^{2}\ln \left( x \right) -4 \right) {x}^{2\,a-1}+{x}^{a-1} \left( \left( {a}^{2}+a \right) \ln \left( x \right) +3\,a+2 \right) \right) a}{ \left( 1- {x}^{a} \right) ^{4}}} \, .$$ Since $\frac{(1-x^a)^4}{a \, x^{a-1}}>0$ we can multiply by this and obtain the objective $$\left( a\left( a-1 \right) {x}^{2\,a}+4\,{a}^{2}{x}^{a}+{a}^{2}+a \right) \ln \left( x \right) + \left( -3\,a+2 \right) {x}^{2\,a}-4 \,{x}^{a}+3\,a+2 \stackrel{!}{\leq} 0 \, .\tag{0}$$
We now need to get rid of the logarithm in order to get a manifest signature. I leave it to you to check that this expression goes to $-\infty$ as $x\rightarrow 0$, while it vanishes at $x=1$. Taking the derivative with respect to $x$ and dividing the resulting expression by $a \, x^{a-1}>0$ yields $$2\,a \left( \left( a-1 \right) {x}^{a}+2\,a \right) \ln \left( x \right) + \left( a+1 \right) {x}^{-a}+ \left( -5\,a+3 \right) {x}^{a} +4\,a-4 \stackrel{!}{\geq}0 \tag{1} \, .$$
Now rinse and repeat. Check the limiting cases (they are $+\infty$ and $0$), derive by $x$ and divide by $a\,x^{a-1}$ to obtain $$2a\left( a-1 \right) \ln \left( x \right) - \left( a+1 \right) {x}^{-2\,a}+4\,{x}^{-a}a-3\,a+1 \stackrel{!}{\leq} 0 \, .\tag{2}$$ Almost there; The limiting cases are manifestly $-\infty$ and $0$ respectively and so after deriving with respect to $x$ and multiplying by $\frac{x^{2a+1}}{2a}>0$ this becomes $$\left( a-1 \right) {x}^{2\,a}-2\,a\,{x}^{a}+a+1 \stackrel{!}{\geq} 0 \, .\tag{3}$$ One more time: At $x=0$ this is $a+1>0$ and at $x=1$ it vanishes again. Deriving with respect to $x$ and dividing by $2a\, x^{a-1}>0$ we have $$(a-1)x^a - a \stackrel{!}{\leq} 0 \, .\tag{4}$$