Well, I think so, and it is less or more equivalent to the basis exchange theorem (which I reinvented answering your question, but looking for a reference to the matroid union theorem I realized that this is a known application of matroid union.)

At first, we ignore that $A$ and $B$ are disjoint (it can not be important a priori, since we may always add twins to our matroid $M$.) But we require that $A\setminus A_i$ and $B_i$ are disjoint.

At second, we suppose that $A$, $B$ are bases (this is rather matter of taste, for dealing with equalities of sizes instead of inequalities). This may be assumed, since we can extend both $A$, $B$ so that they become bases (the partition of $B$ gets a new part $B_{k+1}$) and find the corresponding partition of the extended $A$, it induces a necessary partition of the old $A$. So, from now on $A,B$ are bases and we look for a partition of $A$ such that $(A\setminus A_i)\sqcup B_i$ are bases of our matroid $M$ (then $|A_i|=|B_i|$ is granted.)

The $k=2$ case is the basis exchange theorem which we reproduce here as

**Lemma.** Let $M$ be a matroid.
Let $B$, $A$ be two bases of $M$. Assume that base $B$ is partitioned as $B=B_1\sqcup B_2$. Then base $A$ has such a partition $A=A_1\sqcup A_2$ that both $A_1\sqcup B_2$, $A_2\sqcup B_1$ are bases of $M$.

Now we assume that $k>2$ and the claim is proved for $k-1$. Using lemma, find a partition $A=A_0\sqcup A_k$ such that $A_0\sqcup B_k$ and $A_k\sqcup B_0$
(where $B_0=:B_1\sqcup B_2\ldots \sqcup B_{k-1}$) are both bases of $M$. Consider the matroid $M_0=M/A_k$ on $E\setminus A_k$. Then both $A_0$ and $B_0$ are bases of $M_0$, and by induction hypothesis there exists a necessary partition $A_0=A_1\sqcup \ldots \sqcup A_{k-1}$. I claim that the partition $A=A_1\sqcup \ldots \sqcup A_{k-1}\sqcup A_k$ works. Indeed, for $i=k$ we already know that $(A\setminus A_i)\sqcup B_i$ is a base of $M$. If $i<k$, we know that $(A_0\setminus A_i)\sqcup B_i$ is a base of $M_0=M/A_k$ that yields that $((A_0\setminus A_i)\sqcup B_i)\sqcup A_k$ is a base of $M$, but $((A_0\setminus A_i)\sqcup B_i)\sqcup A_k=(A\setminus A_i)\sqcup B_i$.