As pointed out by Konrad, this follows from the generalisation of van Est's theorem to the continuous case (due to Hochschild and Mostow); namely, that
$$
H_c^m(G,M) \cong H^m(\mathfrak{g},\mathfrak{k};M)
$$
where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{k}$ is the Lie algebra of the maximal compact subgroup of $G$.

For the case in question, $m=1$, $\mathfrak{g} = \mathfrak{so}(n-1,1)$ and $\mathfrak{k}=\mathfrak{so}(n-1)$. I will take $n>2$.

According to Chevalley and Eilenberg, the cohomology $H^m(\mathfrak{g},\mathfrak{k};M)$ is computed from 'horizontal' 'equivariant' cochains in $C^m(\mathfrak{g},M)$, where 'horizontal' means that the cochain vanishes whenever any of its entries belongs to $\mathfrak{k}$ and 'equivariant' means with respect to the action of $\mathfrak{k}$.

Now for the algebras in question, $\mathfrak{g}$ breaks up as $\mathfrak{k} \oplus V$ under the action of $\mathfrak{k}$, where $V$ is the fundamental vector representation of $\mathfrak{k}$, whereas $M = V \oplus \mathbb{R}$, with $\mathbb{R}$ the trivial one-dimensional representation.

Since
$$
C^0(\mathfrak{g},\mathfrak{k};M) = M^{\mathfrak{k}}
$$
it follows that
$$
\dim C^0(\mathfrak{g},\mathfrak{k};M) = 1~.
$$
The differential $\delta: C^0 \to C^1$ is injective, since if $T \in M^{\mathfrak{k}}$ ($T$ is 'timelike' hence the notation)
$$
\delta T(X) = X \cdot T
$$
which does not vanish identically.

On the other hand,
$$
C^1(\mathfrak{g},\mathfrak{k};M) = \text{Hom}(V,M)^{\mathfrak{k}}
$$
is again one-dimensional, hence $C^1 = \delta C^0$ and hence $H^1 =0$.