Rearranging your progress, we get that we need
$$\frac{t^p+1}2>\left(\frac{t-1}{\ln t}\right)^p$$
for all $t>1$.
First, we'll get some necessary conditions on $p$. Let $t=1+\varepsilon$. We have
$$\frac{t^p+1}2=\frac{(1+\varepsilon)^p+1}2=\frac{1+p\varepsilon+\frac{p(p-1)}2\varepsilon^2+O(\varepsilon^3)+1}2=1+\frac p2\varepsilon+\frac{p(p-1)}4\varepsilon^2+O(\varepsilon^3)$$
and
$$\left(\frac{t-1}{\ln t}\right)^p=\left(\frac{\varepsilon}{\ln(1+\varepsilon)}\right)^p=\left(1+\frac{\varepsilon}2-\frac{\varepsilon^2}{12}+O(\varepsilon^3)\right)^p=1+\frac p2\varepsilon+\frac{p(3p-5)}{24}\varepsilon^2+O(\varepsilon^3).$$
So, for $t$ very close to $1$, we need $\frac{p(p-1)}4\geq\frac{p(3p-5)}{24}$. This rearranges to $3p^2\geq p$, which holds exactly when $p\not\in(0,1/3)$. Therefore, if $p\in(0,1/3)$, the inequality does not hold. When $p=0$, we always have $a^p+b^p=2$, so strict inequality fails.
Now we show that, if $p<0$ or $p\geq 1/3$, the inequality holds. For $p<0$, this is not too hard: by a computation in an earlier answer of mine to a similar question, $ab<1$, and so when $p<0$ we have
$$a^p+b^p\geq a^p+\frac1{a^p}\geq 2$$
by the AM-GM inequality.
Now we show the result for $p\geq 1/3$. By the power mean inequality, we have that $\left(\frac{a^p+b^p}2\right)^{1/p}$ is an increasing function of $p>0$, so it suffices to show the result for $p=1/3$.
The name of the game is to reduce the inequality, somehow, to a polynomial inequality. Letting $a=a(t)=\frac{\ln t}{t-1}$ and $b=b(t)=\frac{t\ln t}{t-1}$, we must show that $f(t):=a(t)^{1/3}+b(t)^{1/3}$ strictly exceeds $2$ for all $t>1$. We have $\lim_{t\to 1}f(t)=2$, and so it suffices to show that $f'(t)>0$ for all $t>1$. We know $b$ is increasing in $t$ and that $a-\ln a=b-\ln b$, so $\frac{da}{db}=\frac{1-1/b}{1-1/a}$. We compute
\begin{align*}
\frac{f'(t)}{b'(t)}
&=\frac d{db}\left(a^{1/3}+b^{1/3}\right)\\
&=\frac13b^{-2/3}+\frac13a^{-2/3}\frac{da}{db}\\
&=\frac13b^{-2/3}+\frac13a^{-2/3}\frac{1-1/b}{1-1/a}\\
&=\frac{a^{1/3}}{3b(1-a)}\left(\left(\frac ba\right)^{1/3}(1-a)+(1-b)\right).
\end{align*}
So it suffices to show that $t^{1/3}(1-a)+(1-b)>0$ for $t>1$. We rearrange this as
$$t^{1/3}+1\geq t^{1/3}a+b=\frac{(t^{1/3}+t)\ln t}{t-1}.$$
Let $x=t^{1/3}$; we must show that $\frac{(x+1)(x^3-1)}{x^3+x}\geq 3\ln x$ for $x>1$. Let $g(x):=\frac{(x+1)(x^3-1)}{x^3+x}-3\ln x$. We have $g(1)=0$, so it's enough to show that $g'(x)>0$ for all $x>1$. Indeed,
$$g'(x)=\frac{x^6+3x^4+4x^3+3x^2+1}{x^2(x^2+1)^2}-\frac3x=\frac{(x-1)^4(x^2+x+1)}{x^2(x^2+1)^2}>0.$$
