Set $H^\ast(-) := H^\ast(-,k)$ and $S_a = Sym(a)$. Let $G$ be the wreath product that fits into the extension $$ 1 \to S_a^b \to G \to S_b \to 1.$$
Claim: $H^n(G) = 0$ for $1 \le n \le p-3.$
Proof: The LHS spectral sequence corresponding to the extension is $$E_2^{pq} = H^p(S_b, (H^\ast(S_a)^{\otimes b})^q)$$ where $$(H^\ast(S_a)^{\otimes b})^q = \oplus_{i_1 + ... + i_b = q} H^{i_1}(S_a) \otimes \cdots \otimes H^{i_b}(S_a).$$
Let $1 \le q \le p-3$. Then $i_j \le p-3$ for $j=1,...,b$ and not all $i_j$ can be zero. Hence $H^i(S_a)=0$ for $1 \le i \le p-3$ implies $(H^\ast(S_a)^{\otimes b})^q = 0$.
Thus $E_2^{\ast,q}=0$ for $1 \le q \le p-3$ and $E_2^{p,0} = 0$ for $1 \le p \le p-3$. This shows $H^n(G) = 0$ for $1 \le n \le p-3$.
Remark
Remarks: 1) Even more is true for wreath products: $$H^\ast(G) \cong H^\ast(S_b,(H^\ast(S_a)^{\otimes b})$$ as graded rings (cf. Nakaoka: Homology of the Infinite Symmetric Group, Ann. of. Math. 73(1961),229-257, Theorem 3.3).
2) Since the extension splits, $H^\ast(S_b)$ is a direct summand of $H^\ast(G)$. Hence the vanishing range for the cohomology of $H^\ast(G)$ stated above cannot be improved.
