Note: This is the third version of my answer, one that is, I hope, considerably clearer and cleaner than the previous two.
This can be reduced to a standard problem in real arithmetic, one that, in principle, is solvable, but just how nice the solution will be is a matter of taste, I think.
For simplicity of notation, let me set $\rho_i = 1/t_i^2$ and $\lambda_i = -c_i/{t_i}^2$ for $i=1,2,3$. Also, set $\rho_0 = (c_1/t_1)^2+(c_2/t_2)^2+(c_3/t_3)^2-1$. Consider the following symmetric matrix depending on a parameter $\nu$
$$
Q(\nu) =
\begin{pmatrix}
\rho_0{+}\nu & \lambda_1 & \lambda_2 &\lambda_3\\
\lambda_1 & \rho_1{-}\nu & 0 & 0\\
\lambda_2 & 0 &\rho_2{-}\nu & 0\\
\lambda_3 & 0 & 0 &\rho_3{-}\nu
\end{pmatrix}
$$
and set $D(\nu) = \det\bigl(Q(\nu)\bigr)$.
Claims: (1) The ellipsoid $E$ is disjoint from the unit sphere if and only if there is a value of $\nu$ such that $Q(\nu)$ is either positive or negative definite. (2) The ellipsoid $E$ lies in the interior of the unit ball if and only if the quartic equation $D(\nu)=0$ has $4$ real roots $\nu_i$ that satisfy $0<\nu_0<\nu_1\le\nu_2\le\nu_3$ and, moreover, $Q(\nu)$ is positive definite for some (and, hence, all) $\nu$ satisfying $\nu_0<\nu<\nu_1$.
Note that there are well-established tests for when a matrix is positive definite and for when a quartic polynomial has real roots that are positive. How much you know about the constants $\rho_i$ and $\lambda_i$ will determine how hard actually carrying these tests out will be.
Here is the argument for the claims:
Consider the following two quadratic forms on $\mathbb{R}^4$,
$$
q = -{x_0}^2+{x_1}^2+{x_2}^2+{x_3}^2
$$
and
$$
p = \rho_0\,{x_0}^2 + 2\lambda_1\,x_0x_1+2\lambda_2\,x_0x_2+2\lambda_3\,x_0x_3
+ \rho_1\,{x_1}^2+ \rho_2\,{x_2}^2+ \rho_3\,{x_3}^2,
$$
To know whether or not the ellipsoid $E$ and the unit sphere have a real intersection point is the same as knowing whether there is a nonzero vector in $\mathbb{R}^4$ that is a null vector for both $q$ and $p$. Now, it is a standard fact of linear algebra that, on a real vector space of dimension greater than $2$, a pair of quadratic forms has a positive definite linear combination if and only if they have no common null vector other than the zero vector. (See the Footnote for a proof of this standard fact.) This establishes the first claim.
The second claim depends on the first: If $E$ (assumed to have nonempty interior) lies in the interior of the unit ball, then $p$ and $q$ have no common null vector, so, by the first claim, there is some value of $\nu$ for which $Q(\nu)$ is definite (either positive or negative). It then follows from the usual linear algebra proofs that $p$ and $q$ can be simultaneously diagonalized, i.e., (since $q$ and $p$ clearly have type $(3,1)$ and $(1,3)$ respectively), there is a basis of $\mathbb{R}^4$ such that, in the corresponding coordinates $y_i$, we have
$$
q= -{y_0}^2+{y_1}^2+{y_2}^2+{y_3}^2
$$
and
$$
p = -\nu_0\,{y_0}^2+\nu_1\,{y_1}^2+\nu_2\,{y_2}^2+\nu_3\,{y_3}^2
$$
for some (nonzero, since $p$ is nondegenerate) numbers $\nu_i$. Since $E$ is contained in the interior of the unit ball if and only if the closure of the negative cone of $p$ is contained in the negative cone of $q$, it follows that all of the $\nu_i$ must be positive, and, in fact, rearranging $\nu_1,\nu_2,\nu_3$ if necessary, we must have $0<\nu_0<\nu_1\le \nu_2\le \nu_3$. It is now clear that the roots of $D(\nu)=0$ are the $\nu_i$ and that $Q(\nu)$ is positive definite when $\nu_0<\nu<\nu_1$, since $p-\nu q$ is.
Conversely, if $Q(\nu)$ and $D(\nu)$ satisfy the hypotheses, then $p-\nu q$ is a positive definite quadratic form and so the negative cone of $q$ must contain the negative cone of $p$, which implies that $E$ lies in the interior of the unit ball.
Footnote added on 9/1/13: To prove the Standard Fact:
One direction is obvious: If the two quadratic forms have a positive definite linear combination, then they have no common null vector.
For the other direction, let $p$ and $q$ be quadratic forms on $\mathbb{R}^n$ (where $n>2$) that have no common zero other than $0\in\mathbb{R}^n$. We need to show that some linear combination of $p$ and $q$ is positive definite.
Consider the map $f = (p,q):\mathbb{R}^n\to\mathbb{R}^2$, which, by hypothesis, sends only the origin to the origin. The normalized map
$$
F(v) = \frac{f(v)}{|f(v)|}
$$
is well-defined and smooth on $\mathbb{R}^n$ minus the origin and is even, i.e., $F(v)=F(-v)$, and homogeneous of degree $0$, so it induces a well-defined smooth map $\phi:\mathbb{RP}^{n-1}\to S^1\subset\mathbb{R}^2$.
Now, the image of $S^{n-1}\subset\mathbb{R}^n$ under $f$ lies in an open halfspace in $\mathbb{R}^2$ if and only if some linear combination of $p$ and $q$ is positive definite, so suppose that this (connected) image does not lie in any open halfspace. Then there will exist two nonzero vectors $x,y\in\mathbb{R}^n$ such that $f(x) = - f(y)\not=(0,0)$. Then, by the usual polarization identity for quadratic forms, one has
$$
f(\cos\theta\,x+\sin\theta\,y)
= \cos2\theta\, f(x) + \sin2\theta\, \tfrac12f(x{+}y)\ \ (\not=0\ \text{for all $\theta$}).
$$
This implies that the path $\gamma(t) = [\cos\theta\,x+\sin\theta\,y]$ for $0\le t\le \pi$, which is a closed path in $\mathbb{RP}^{n-1}$ that generates $H_1(\mathbb{RP}^{n-1},\mathbb{Z})\simeq \mathbb{Z}_2$, is mapped by $\phi$ to a generator of $H_1(S^1,\mathbb{Z})\simeq\mathbb{Z}$, which is absurd.
Note that the proof breaks down for $n=2$ because $H_1(\mathbb{RP}^{1},\mathbb{Z})\simeq \mathbb{Z}$ instead of $\mathbb{Z}_2$. This is good because the pair of quadratic forms $p = x^2-y^2$ and $q = 2xy$ on $\mathbb{R}^2$ have no common null vector and yet have no positive definite linear combination.
