I believe we can. Let $\ell_n \to \ell$ in your topology, and fix $t_0 \in \mathbb R_+$. We show that $\Gamma(\ell_n)(t_0) \to \Gamma(\ell)(t_0)$. We work over the interval $[0, T]$ with $T > t_0$.

**Step 1:** We first note that $\ell_n$ converges to $\ell$ in measure.

Indeed, let $\varepsilon > 0$ be arbitrary. As $\ell$ is nonincreasing, $\ell$ contains only jump discontinuities and for any $n > 0$, there exist only finitely many jumps with magnitude greater than $\frac{\varepsilon}{2^j}$. Cover these jumps with finitely many closed intervals $C_i^j$ of total length less than $\frac{\varepsilon}{2^j}$. Divide the complement into intervals $D_i^j$ on which $\ell$ varies by no more than $\frac{\varepsilon}{2^j}$, and consider the partition $\mathcal P_j ;= C_i^j \cup D_i^j$. Applying the pointwise convergence of $\ell_n$ to $\ell$ near the endpoints of the $C_i^n$ and the monotonicity of $\ell_n$ and $\ell$ now allows us to conclude.

**Step 2:** We show that $X^{\ell_n} \to X^{\ell}$ uniformly in probability.

Namely, that $$P(\sup_{s \in [0, T]} | X_t^{\ell_n} - X_t^{\ell}|
\geq \frac{1}{2^k}) \to 0.$$

To see this, note that we can write

$$X^{\ell}_t:=1+t+\int_0^t\frac{1}{1+\ell(s)} + g_n(s) \ dW_s \, \quad \forall t\ge 0.$$

with $g_n := \frac{1}{1 + \ell_n(s)} - \frac{1}{1 + \ell(s)} \to 0$ in $L^2$. Whereby the aforementioned convergence follows from the convergence in measure of $\ell_n$ to $\ell$.

Note that $X_s^{\ell_n} - X_s^{\ell}$ is a martingale, and that $X_T^{\ell_n} - X_T^{\ell}$ is Gaussian with mean $0$, and variance $||g_n||_{L^2}$.

And so by Doob’s (sub)martingale inequality and symmetry of the Gaussian distribution, we have

$$P(\sup_{s \in [0, T]} | X_s^{\ell_n} - X_s^{\ell}|
\geq \frac{1}{2^k}) \leq 2^{k+1} E[(X_T^{\ell_n} - X_T^{\ell})^+] \to 0,$$

as $n \to \infty$.

**Step 3:** Conclusion.

Let $\varepsilon > 0$ be arbitrary. We note that by continuity of $X_n^{\ell}$ we can write the event $\{\tau^{\ell} > t\}$ as $$\bigcup_{k \in \mathbb N} A_k := \bigcup_{k \in \mathbb N} \{X_s \geq \frac{1}{k}, \ \forall s \in [0,t_0]\}.$$

By continuity from below, we have that for some $k_0 \in \mathbb N$ that $P(A_{k_0}) > P(\{\tau^{\ell} > t\}) - \frac{\varepsilon}{2}$.

Since $X^{\ell_n} \to X^{\ell}$ uniformly in probability, for all large enough $n$, we have

$$P(\sup_{s \in [0, T]} | X_s^{\ell_n} - X_s^{\ell}|
\geq \frac{1}{2k_0}) < \frac{\varepsilon}{2},$$

so that for all $n > N$ we have

$$\Gamma(\ell_n)(t_0) = P(\{\tau^{\ell_n} > t_0\}) > P(A_{k_0} \cup \{\sup_{s \in [0, T]} | X_s^{\ell_n} - X_s^{\ell}| > \frac{1}{2k_0} \}^c) > P(\{\tau^{\ell} > t_0\}) - \varepsilon$$

For the reverse inequality, we argue similarly - we note that, up to $P$-null sets, we can write the event $\{\tau^{\ell} \leq t_0\}$ as

$$\bigcup_{i \in \mathbb N} \{\inf_{s \in [0, t_0)}\ X_s^{\ell} \leq -\frac{1}{i}\},$$

In order for the above representation to hold, we need to show that $\{\inf_{0\le s\le t}X^{\ell}_s\le 0\}=\{\inf_{0\le s\le t}X^{\ell}_s< 0\}$, up to a $P$-null set.

For this, it suffices to use Lévy's characterization. More precisely, we can write $X^{\ell}_t=1+t+B_{\langle X^{\ell}\rangle_t}\equiv 1+t+B_{L(t)}$, where $B$ denotes a Brownian motion and $L(t):=\int_0^t ds/(1+\ell(s))^2$. Therefore

$$\inf_{0\le s\le t}X^{\ell}_t = \inf_{0\le u\le L(t)}\{1+L^{-1}(u)+B_t\},$$

which implies the desired result as $\inf_{0\le u\le L(t)}\{1+L^{-1}(u)+B_t\}$ admits a density.

Thus with a similar calculation, we obtain $$\Gamma(\ell_n)(t) < P(\{\tau^{\ell} < t_0\}) + \varepsilon$$

for all large enough $n$.

Since $\varepsilon$ was arbitrary, we conclude that $\Gamma(\ell_n)(t_0) \to \Gamma(\ell)(t_0)$, as was to be shown.