= Solution
For a fixed environment, use two <independent> copies of the walk to express $Z_t^2=\langle\kappa_t^{(1)}\kappa_t^{(2)}\rangle$. The product is finite and nonnegative, so we can interchange the two expectations. At each time $j$, if the walks occupy different sites, the two environment signs are <independent> and their factors have product expectation $1$. If they coincide, both factors use the same sign and
$$
\mathbb E(1+\varepsilon h)^2=1+\varepsilon^2.
$$
Different time layers are <independent> even when a site is visited again. Therefore, with the simultaneous collision count $I_t=\sum_{j=1}^t\mathbf1_{\{\xi_j^{(1)}=\xi_j^{(2)}\}}$,
$$
\boxed{\mathbb EZ_t^2=\left\langle(1+\varepsilon^2)^{I_t}\right\rangle.}
$$
This is the <collision-count second moment for independent path weights>. It involves equal-time intersections, not arbitrary intersections of the two spatial ranges.
The difference walk $D_j=\xi_j^{(1)}-\xi_j^{(2)}$ has <independent> bounded increments symmetric about zero and a full-dimensional step distribution. It may hold at zero for one step; such a hold counts as a positive-time return. The allowed transience fact gives
$$
\rho=\mathbb P_0(D_j=0\text{ for some }j\ge1)<1.
$$
Moreover $\rho>0$ since the two first steps can be equal. Restarting the <independent increments> after every return to zero shows that the total positive-time return count $I_\infty$ has the zero-based <geometric distribution>
$$
\mathbb P(I_\infty=k)=(1-\rho)\rho^k,\quad k=0,1,2,\ldots.
$$
Indeed the probability of at least $k$ returns is $\rho^k$, and subtracting successive tails gives this law. Consequently, whenever $\rho(1+\varepsilon^2)<1$,
$$
\sup_t\mathbb EZ_t^2\le\left\langle(1+\varepsilon^2)^{I_\infty}\right\rangle
=\sum_{k=0}^\infty(1-\rho)[\rho(1+\varepsilon^2)]^k
=\frac{1-\rho}{1-\rho(1+\varepsilon^2)}<\infty.
$$
A sufficient small-noise condition is $0<\varepsilon<\min(1,\sqrt{(1-\rho)/\rho})$.
The <L2 martingale convergence theorem> now gives $Z_t\to\zeta$ in $L^2$ as well as almost surely. One can see the <mean> preservation directly: for $t\ge s$, the <martingale> identity gives $\mathbb E(Z_t-Z_s)^2=\mathbb EZ_t^2-\mathbb EZ_s^2$. The <second moments> are increasing and bounded, so the sequence is Cauchy in $L^2$ and its limit is the already identified almost-sure limit. Thus
$$
\boxed{\mathbb E\zeta=\lim_t\mathbb EZ_t=1.}
$$
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