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 $H^1 =0$.
