The trajectory is defined for $t \ge 0$ and $t \neq 1$. Let $I= [0,1) \cup(1, \infty)$.
$x$ is a strictly increasing continuous bijection $[0,1) \to [0,\infty)$ and a strictly increasing continuous bijection
$(1,\infty) \to (-\infty,0)$. Hence $(x(t), y(t))$, $t \in I$ defines a function $f:\mathbb{R} \to \mathbb{R}$.
It is straightforward to check that the maps $x:[0,1) \to [0,\infty)$ and $x:(1,\infty) \to (-\infty,0)$ are actually homeomorphisms.
It is straightforward to check that $\lim_{s \to \infty} \phi(s) = L$ iff $\lim_{t \uparrow 1} \phi(x(t)) = L$.
Then we have $\lim_{s \to \infty} {f(s) \over s} = \lim_{t \uparrow 1} {f(x(t)) \over x(t)} = \lim_{t \uparrow 1} {y(t) \over x(t)} = \lim_{t \uparrow 1} (1-2t^2) =-1$, and similarly,
$\lim_{s \to \infty} {f(s) + s} = \lim_{t \uparrow 1} {f(x(t)) + x(t)} = \lim_{t \uparrow 1} {y(t)+ x(t)} = \lim_{t \uparrow 1} 2t =2$.
Note that it is also true that $\lim_{t \to 1} {y(t) \over x(t)} = \lim_{t \to1} (1-2t^2) =-1$ and $\lim_{t \to 1} {y(t)+ x(t)} = \lim_{t \to 1} 2t =2$, and using the
$x:(1,\infty) \to (-\infty,0)$ portion of the curve, we can show in a similar manner that
$\lim_{s \to -\infty} {f(s) \over s} = -1$ and
$\lim_{s \to -\infty} {f(s) + s} = 2$.
Elaboration
Take the case where $L$ is finite for example. Also, note that I am dealing with $x:[0,1) \to [0,\infty)$ here. Note that $x$ is strictly increasing on this domain:
Suppose for all $\epsilon>0$ there is some $S$ such that if $s > S$ then $|\phi(s)-L| < \epsilon$. Now let $t_0= x^{-1}(S), \delta = 1-t_0$ and note that if $0< |1-t| < \delta$ (and implicitly we have $t \in [0,1)$) we have $|\phi(x(t))-L| < \epsilon$.
The other direction is similar, suppose we have some $\delta>0$ such that
if $0< |1-t| < \delta$ (and implicitly $t \in [0,1)$) then $|\phi(x(t))-L| < \epsilon$. We can assume that $\delta<1$. Let $S=x(1-\delta)$, then if $s > S$ there is some $t \in (1-\delta,1)$ such that $x(t)=s$ and so
$|\phi(s)-L| = |\phi(x(t))-L| < \epsilon$.
The case for infinite $L$ is similar.
Best Answer
The second item of your definition of an oblique asymptote entails that you should get closer to the oblique line upon just panning to the right, without having to de-zoom.
This is indeed not the case here although it would be very hard to observe since, as others have pointed out in comments, $\ln$ goes to infinity very slowly. A more striking example would be $x+\sqrt{x}$ maybe.
The weaker property that the curve looks more and more like an oblique line upon dezooming corresponds to only the first item in your definition, which, as you have shown, indeed is the case for $x+\ln(x)$.