I'm going to assume that $0 \preceq A$, that is, that $A$ is itself positive semidefinite, as the result fails for more general $A$ (and the thing defined as the "ellipsoid" of $A$ is not a geometrical ellipsoid; e.g. if $A$ is the diagonal $2 \times 2$ matrix with diagonal $(-1, 0)$ and $B$ is the $2 \times 2$ identity matrix, we do have $A \preceq B$ in the sense that $B - A$ is positive semidefinite, but the "ellipsoid" of $A$ is an unbounded set whose boundary is the hyperbola $\{(x,y) \in \mathbb{R}^2: y^2 - x^2 = 1\}$ and this is not contained in the ellipsoid of $B$, aka the unit disc in $\mathbb{R}^2$).
Once we assume that $A$ is positive semidefinite it follows from $A \preceq B$ that $B = (B-A) + A$ is also positive semidefinite (since a sum of PSD matrices is PSD).
Since the definition of ellipsoid involves the inverse we should also assume that $A$ is invertible (and this implies that $B$ is--- e.g. because it implies the existence of a positive number $c$ with $cI \preceq A$ and transitivity of $\preceq$ implies then that $cI \preceq B$).
Now it is well known (but nontrivial) that $0 \preceq A \preceq B$ and $A$ invertible together imply that $0 \preceq B^{-1} \preceq A^{-1}$. [The easiest way to prove this depends sensitively on how "positive semidefinite" is defined, and there are various approaches to this. A functional analysis book, for example, would define the concept in a way that does not reference eigenvalues, since the notion makes perfectly good sense for operators that don't have any.] Look it up your favorite reference or find a proof that you like online.
Anyhow: once you know the given hypotheses imply that $B^{-1} \preceq A^{-1}$ it is simple. For any vector $x$ you have from this operator relation that $x^t (A^{-1} - B^{-1}) x \geq 0$, which implies (by expanding the matrix product and moving one term over to the other side) that for any $x$ $ x^t A^{-1} x \geq x^t B^{-1} x; $ thus if the left hand side is $\leq 1$, so is the right hand side, thus the ellipsoid of $A$ is contained in that of $B$.
For the converse: if it is not the case that $A \preceq B$ (under the standing hypotheses that $A, B$ are both PSD and invertible) then it cannot be the case that $B^{-1} \preceq A^{-1}$ (or we could deduce from this that $A \preceq B$ by the well known but nontrivial fact I mentioned earlier, with the roles of $A$ and $B$ now played by $B^{-1}$ and $A^{-1}$), so there must be a vector $x_0$ with $x_0^t(A^{-1} - B^{-1}) x_0 < 0$, and for this vector we have $ x_0^t B^{-1} x_0 > x_0^t A^{-1} x_0. $ Since $A^{-1}$ is invertible, the right hand side must be positive, so by scaling $x_0$ as necessary we can make the right hand side equal to $1$. The inequality then shows us that the vector $x_0$ is in the ellipsoid of $A$, but not in that of $B$.