= Solution
The <intensity function of a point process> is positive and continuous inside the interval, so its <measure> is finite on every <compact> subinterval. Consequently the <Poisson point process> is locally finite there. Its nonatomic intensity makes it simple and gives no point at $0$ <almost surely>. At the endpoints,
$$
\lambda(x)\sim\frac{1}{4(1-x)^3}\quad(x\uparrow1),\qquad\lambda(x)\sim\frac{1}{8(1+x)^2}\quad(x\downarrow-1).
$$
Both half-intervals therefore have infinite intensity. More explicitly, exhaust either half by increasing <compact> intervals with means $t_j\to\infty$. For any fixed integer $k$,
$$
\mathbb P\{\#\Pi\text{ in that half}\leq k\}\leq e^{-t_j}\sum_{i=0}^k\frac{t_j^i}{i!}\longrightarrow0.
$$
Taking the <countable> union over $k$ proves \b[infinitely many points on each side <almost surely>].
Local finiteness excludes accumulation in the interior, including at $0$. Hence there is a largest negative point and a smallest positive point. For example, given any negative point, the <compact> interval between it and $0$ contains finitely many points and therefore a largest one. Call it $X_0$; call the smallest positive point $X_1$, and enumerate consecutively in both directions. Infinite counts give every integer index, and the lack of interior accumulation implies $X_n\uparrow1$ as $n\to\infty$ and $X_n\downarrow-1$ as $n\to-\infty$. Thus the specified two-sided labeling exists <almost surely>.
Use the cumulative-intensity transformation
$$
\boxed{f(x)=\int_0^x\frac{dt}{(1+t)^2(1-t)^3}.}
$$
It has $f(0)=0$, positive derivative, and limits $-\infty,+\infty$ at the two endpoints, so it is an increasing <bijection> onto $\mathbb R$. One explicit expression is
$$
f(x)=\frac{3}{16}\log\frac{1+x}{1-x}-\frac{1}{8(1+x)}+\frac{1}{4(1-x)}+\frac{1}{8(1-x)^2}-\frac14.
$$
For any bounded interval $(a,b)$ in the target, its preimage has intensity
$$
\int_{f^{-1}(a)}^{f^{-1}(b)}\lambda(x)\,dx=b-a.
$$
Disjoint target sets have disjoint preimages, so their counts are <independent>. This proves directly, or by the <Poisson mapping theorem>, that the transformed points form a unit-rate <Poisson process> on the real line.
Put $Y_n=f(X_n)$. On the positive half-line, the successive spacings are <independent> <exponential random variables> of <mean> $1$, so $Y_n=E_1+\cdots+E_n$. The <strong law of large numbers> gives $Y_n/n\to1$ <almost surely>. On the negative half-line, independently, the reflected points satisfy $-Y_{-k}=E'_1+\cdots+E'_{k+1}$, because $Y_0$ is the first point to the left of zero. Thus $-Y_{-k}/k\to1$, or $Y_n/n\to1$ also as $n\to-\infty$. The shift by one has no effect on these limits.
From the explicit formula, or by integrating the endpoint intensity asymptotics,
$$
f(x)(1-x)^2\longrightarrow\frac18\quad(x\uparrow1),\qquad[-f(x)](1+x)\longrightarrow\frac18\quad(x\downarrow-1).
$$
Substitute $x=X_n$ and use the <strong law for Poisson arrival times>. At the right endpoint,
$$
n(1-X_n)^2=\frac{n}{f(X_n)}\,f(X_n)(1-X_n)^2\longrightarrow\frac18,
$$
so
$$
\boxed{\sqrt{2n}\,(1-X_n)\longrightarrow\frac12\quad(n\to+\infty),\ \text{almost surely}.}
$$
At the left endpoint the corresponding result is
$$
\boxed{|n|(1+X_n)\longrightarrow\frac18\quad(n\to-\infty),\ \text{almost surely}.}
$$
Equivalently $8|n|(1+X_n)\to1$. The different powers of $|n|$ reflect the different endpoint singularity orders, an instance of <power-law boundary accumulation of a Poisson point process>.
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