(This is basically a more detailed version of Eoin's comment.)
I assume that you are considering division algebras over a field $k$, i.e., $Z(A) = k$. If $B$ is a subalgebra of dimension $2$ of $A$, then $B$ is always commutative, and since it lives inside the division algebra $A$, it must be a quadratic field extension of $k$.
If $B/k$ is a separable extension, then it admits a non-trivial automorphism $\sigma$. The Skolem-Noether theorem tells us that every automorphism of a simple subalgebra extends to an inner automorphism of the whole algebra. In particular, $\sigma$ extends to an inner automorphism of $A$, say conjugation by $a \in A$, and of course we must have $a \not\in B$ since $\sigma$ is non-trivial.
If $B/k$ is an inseparable extension, then we have $\operatorname{char}(k)=2$ and $\operatorname{Gal}(B/k)=1$. So any inner automorphism of $A$ normalizing $B$ must, in fact, centralize $B$. Now let $C = C_A(B)$ be this centralizer. Then, by the centralizer formula, we have $\dim_K(A) = \dim_K(B) \dim_K(C)$. In particular, if $\dim_K(A) = 4$ (the smallest possible since the dimension is always a square), we have $C = B$, and this provides the kind of examples you are looking for. On the other hand, if $\dim_K(A) > 4$, then $\dim_K(C) > 2$ so $C$ strictly contains $B$. In other words, there are elements of $C \setminus B$ that conjugate $B$ to itself, so there are no new examples in this situation.
(For a reference, both the Skolem-Noether theorem and the centralizer formula can be found in section 8.4 of W. Scharlau's book "Quadratic and hermitian forms" from 1985.)
