I don't think this is a trivial fact but maybe it is something that follows without too much difficulty from something you've seen before. I'll offer two ways of seeing this.
The more advanced of the two (but accessible if you happen to know things about connections between Brownian motion and PDE) is to recall that $u(x,t) = \mathbb{P}_x(T_1 > t)$ solves the heat equation on $[-1,1]$ with boundary conditions that can be read off the above formula for $u$. Then one can write $u$ in terms of a Fourier series that has an exponential term and apply brutal estimates. I leave this as a straightforward exercise if you know the PDE connection.
The approach that is more accessible again requires us to consider $u(x,t)$ as above (the idea is that we want to use the strong Markov property of Brownian motion so we will want to be able to consider different starting points for the BM). I claim first that there exists $q \in (0,1)$ such that for $k \in \mathbb{N}$ and $x \in (-1,1)$ one has that $u(x,k) \leq q^k$.
I prove this by induction. First, note that
$$\mathbb{P}_x[T_1 < 1] \geq \mathbb{P}[|B_1 - x| > 2] = \mathbb{P}_0[|B_1| > 2] =: p > 0.$$
Let $q = 1 - p$ so that the above forms the base case for the induction. Then by the strong Markov property and the fact that $B_1 - x$ is distributed as a standard Gaussian under $\mathbb{P}_x$ we have that
\begin{align*}
\mathbb{P}_x[T_1 > k] \leq& (2 \pi)^{-1/2} \int_{-1}^{1} e^{-|x-y|^2/2} \mathbb{P}_y[T \geq k-1] dy \\ \leq& q^{k-1} \mathbb{P}_x(B_1 \in [-1,1]) \\ \leq& q^{k-1} \mathbb{P}_x(|B_1 - x| \leq 2) = q^k
\end{align*}
completing the induction.
Your desired result then follows by taking $q = e^{-b}$ and writing $$\mathbb{P}_0[T_1 > t] \leq \mathbb{P}_0[T_1 > \lfloor t \rfloor] \leq e^{-b \lfloor t \rfloor} = e^{b(t - \lfloor t \rfloor)} e^{-bt} \leq e^b e^{-bt}$$
In fact, this proof extends with basically no work to show that the exit time of a $d$-dimensional Brownian motion of an open bounded set $U$ has exponential tails and hence finite moments of all orders.
