Not quite. Think of the vertices of a square; a tall and narrow ellipse passes through them, but so does a short and wide ellipse. So you don't get uniqueness.
When you (naturally) ask about 5 points, it turns out that there is a unique conic containing any five points...but it's not necessarily an ellipse, even if the points form a nice convex set. Why? Think of 5 points on a parabola. The unique conic that fits these is...that parabola! So there's no ellipse that passes through them.
The projective space of conics is a fascinating introduction to algebraic geometry. The book on Projective Geometry by Pierre Samuel is a pretty nice intro if your other math skills are rock solid, but "other" includes abstract algebra in this case, so it'll be a while before you're ready to read it.
Let me just add another remark here, which hints at the projective geometry thing. Suppose that
H_1(x, y) = Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F
and that the four points $(x_i, y_i)$, $i = 1, 2, 3, 4$, have the property that
H_1(x_i, y_i) = 0 (i = 1, 2, 3, 4).
Then the equation
H_1(x, y) = 0
defines a conic containing the four points.
Now suppose that $H_2$ is another such polynomial (quadratic in $x$ and $y$) and that the equation $H_2(x, y) = 0$ is satisfied by the same four points. (Think of $H_1 = 0$ as defining the red ellipse in @muaddib's answer, and $H_2 = 0$ as defining the blue one.)
Then for any $t$, the polynomial
Q_t(x, y) = (1-t) H_1(x, y) + t H_2(x, y)
also is zero at the four points. So in the "space of all conics", if two conics $C_1$ and $C_2$ pass through four points, so do all conics on the "line between $C_1$ and $C_2$" (i.e., those like $Q_t$ above).
To close the argument:
In general, if you have four points $P_1, P_2, P_3, P_4$, and they lie on an ellipse, $E$, you can pick a 5th point, say $R_0$, not close to any of the $P_i$, that's also on that ellipse. Now you can move $R_0$ very slightly to get a new point $R_1$, and consider the conic $C$ passing through
P_1, P_2, P_3, P_4, R_1.
If $R_1$ is close enough to $R_0$, then the conic $C$ is very close to the original ellipse, and must therefore also be an ellipse (a statement that needs proving, by the way -- it's not at all obvious!).
But now you can take the quadratic for $E$, say $H_1$, and the quadratic for $C$, say $H_2$, and form a combination like $(1-t)H_1 + tH_2$, where $t$ is any real number, to get another conic passing through the four points, so there's a whole infinity of conics passing through these points. By the same argument as above -- the "not at all obvious" one -- infinitely many of these conics must also be ellipses.
In short: no, even in the case where the four points are in "convex position" and there's an ellipse through them, there's always another ellipse through them, and indeed, infinitely many others.