= Solution
Write $c=1/(a-1)$. If $X_0=x>c$, then for either possible sign,
$$
X_{n+1}\geq aX_n-1,\qquad
X_{n+1}-X_n\geq(a-1)X_n-1>0
$$
as long as $X_n>c$. Thus the process stays above $c$, is strictly increasing, and induction gives $X_n\geq c+a^n(x-c)\to\infty$. If $x<-c$, the symmetric argument makes the process strictly decreasing to $-\infty$. This proves the first assertion pathwise, with the strict sufficient threshold $|x|>c$.
For an arbitrary starting point, a proof is needed that the process actually reaches such a region. Fix $L>c$ and choose an integer $m$ so
$$
\sum_{j=0}^{m-1}a^j=\frac{a^m-1}{a-1}>L.
$$
At a block starting at time $km$, if $X_{km}\geq0$, the next $m$ signs all equal to $+1$ give $X_{(k+1)m}>L$. If $X_{km}<0$, all $-1$ signs instead give $X_{(k+1)m}<-L$. Conditional on the current <natural filtration>, the selected block has probability $\varepsilon=2^{-m}$, by <independence> of future signs. If $\tau=\inf\{n:|X_n|>L\}$, iterated <conditional expectation> gives
$$
\mathbb P_x(\tau>km)\leq(1-\varepsilon)^k\longrightarrow0.
$$
This is a <geometric tail bound from a uniform escape probability>, so $\tau<\infty$ <almost surely>. Once outside, the sign is preserved and, for every future noise sequence,
$$
|X_{\tau+j}|\geq c+a^j(|X_\tau|-c)\longrightarrow\infty.
$$
Consequently the <explosive affine recursion with symmetric bounded noise> has
$$
\boxed{\mathbb P_x\!\left(\lim_{n\to\infty}|X_n|=\infty\right)=1\quad\text{for every }x\in\mathbb R.}
$$
The block argument closes the gap between escape from a large initial value and escape from an arbitrary one; no assertion about the absence of atoms in an infinite random series is necessary.
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