Iosif Pinelis has already posted an answer, but here is an alternate answer communicated to me by Yuval Peres on a different website. Any typoes/mistakes are most definitely mine.

Write

$$\tau = \text{min}\{t > 0 \, : \, W_t \geq \frac{1}{\varepsilon}\}.$$

By the reflection principle, we have

$$\mathbb P(\tau \leq 1) = \mathbb P(A_\varepsilon) = 2 \Phi (-\frac{1}{\varepsilon}),$$

where $\Phi(x) := \int_{-\infty}^x (2\pi)^{-\frac{1}{2}} e^{-\frac{t^2}{2}} \, dt$ denotes the CDF of the standard normal distribution.

Using the strong Markov property at time $\tau$, we have that $|W_1 - W_\tau|$ is a standard half normal random variable with parameter $\sigma = 1 - \tau$, independent of $\mathcal F_\tau$.

Thus we compute

$$\mathbb E \big [|W_1 - \frac{1}{\varepsilon}| \, \big | \, \tau \leq 1 \big ]$$
$$= \mathbb E \big [|W_1 - W_\tau| \, \big | \, \tau \leq 1 \big ]$$
$$ = \frac{\mathbb E[\mathbf 1_{\{\tau \leq 1\}} |W_1 - W_\tau|]}{\mathbb P(\tau \leq 1)}$$
$$ = \frac{\mathbb E[\mathbf 1_{\{\tau \leq 1\}} \mathbb E[ |W_1 - W_\tau| \big | \sigma(\tau) ]]}{\mathbb P(\tau \leq 1)}$$
$$ = \frac{\mathbb E \big [\mathbf 1_{\{\tau \leq 1\}} \sqrt{\frac{2}{\pi} (1 - \tau)} \big ]}{\mathbb P(\tau \leq 1)}$$
$$ \leq \sqrt \frac{2}{\pi}.$$

Thus

$$\mathbb E_{\mathbb P^\varepsilon} [|\varepsilon W_1 - 1|] = \mathbb E[|\varepsilon W_1 - 1 | \big | A_\varepsilon] \leq \varepsilon \sqrt{\frac{2}{\pi}}$$

which tends to $0$, as desired.

In fact, the bound above may be further improved as observed by Yuval Peres.

We compute, for any $0 < \delta < 1$,

$$\mathbb P[\tau \leq 1 - \delta \, | \, \tau \leq 1] = \Phi \left ( -\frac{1}{\sqrt{ 1 - \delta} . \varepsilon } \right ) \big /\Phi \left ( -\frac{1}{\varepsilon } \right ) $$

$$\leq \frac{\text{exp} \big (\frac{1}{2 \varepsilon^2} \big )}{\text{exp} \big (\frac{1}{2 \varepsilon^2 (1 - \delta)} \big )}$$|

where the inequality above is due to the fact that $\Phi(-r) e^{\frac{r^2}{2}}$ is decreasing in $r$.

Thus,

$$\mathbb E \left [ \frac{\sqrt{1 - \tau}}{\varepsilon} \, \big | \, \tau \leq 1 \right ] = \int_0^\infty \mathbb P(\sqrt{1 - \tau} \geq r\varepsilon \, | \, \tau \leq 1) \, dr$$

$$ \leq \int_0^\infty \text{exp}\big (-\frac{r^2}{2} \big )\, dr = \sqrt{\frac{\pi}{2}}$$

And so finally,

$$\mathbb E_{P^\varepsilon} [|\varepsilon W_1 -1|] \leq \sqrt{\frac{2}{\pi}}\sqrt{\frac{\pi}{2}} \varepsilon^2 = \varepsilon^2$$

and this is sharp asymptotically.