Let $B$ be a central simple algebra over an algebraically closed field $k$, and let $A\subset B$ be a semi-simple $k$-subalgebra. By this I also mean $k$ is central in $A$ as well, and the map $k\to A\to B$ is the identity map onto the center of $B$.
Now we assume that the map $k\to A$ surjects onto the center of $A$. In particular, not only do $A$ and $B$ have the same center, $k$, but their identities match and the map $A\to B$ is unitary and central (i.e. it sends the center of $A$ into the center of $B$).
For instance, the inclusion $M_n(k)\subset M_m(k)$ for $m>n$ via block matrices does not qualify. For $\sum_{i=1}^Nm_i=m$, the inclusion $$\prod_{i=1}^NM_{m_i}(k)\subseteq M_m(k)$$ does not qualify either. It is a unitary map, but the center of $A$ is $k^N$ while the center of $B$ is $k$.
If $A\neq k$, do we have $A=B$?
Remark As a special case, if we knew that the inclusion of $A$ in $B\simeq M_n(k)=\text{End}_k(k^{\oplus n})$ makes $k^{\oplus n}$ into a simple $A$-module, we could use Burnside's Theorem to conclude $A=B$.
I was guessing the answer is "no", but I can't find an example so I started wondering. I'd appreciate any help very much!
What about this one: take $B=M_4(k)$ and $$ A=\left\{\begin{pmatrix}a&b&0&0\\c&d&0&0\\0&0&a&b\\0&0&c&d\end{pmatrix}\mid a,b,c,d\in k\right\}, $$ i.e. the diagonal embedding of $M_2(k)$ into $M_4(k)$. Then I think $A$ and $B$ are both central and simple with matching centres and units.
Also, I think that $A$ is forced to be simple in your setup, because if it is a product of $n$ central simple algebras then its center will be $k^{\oplus n}$.