Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix, we see that:
$\det \begin{bmatrix} t & -a_0 \\ -1 & t-a_1 \end{bmatrix} = t(t-a_1) - a_0 = t^2 - a_1t - a_0$
and, by expansion along the first row:
$\det \begin{bmatrix} t & 0 & -a_0 \\ -1 & t & -a_1 \\ 0 & -1 & t-a_2 \end{bmatrix} = t \times\det \begin{bmatrix} t & -a_1 \\ -1 & t-a_2 \end{bmatrix} + (-a_0) \det\begin{bmatrix} -1 & t \\ 0 & -1 \end{bmatrix}$
$= t[t(t-a_2) - a_1] - a_0 = t^3 - a_2t^2 - a_1t - a_0 $
So it looks like:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$
Which we can prove by induction.
Assume that:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & t-a_{n-1} \end{bmatrix} = t^{n} - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - ... - a_2t^2 - a_1t - a_0$
Then, by expansion along the first row:
$\det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_0 \\ -1 & t & 0 & \cdots & 0 & -a_1 \\ 0 & -1 & t & \cdots & 0 & -a_2 \\ 0 & 0 & -1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} = t \det \begin{bmatrix} t & 0 & 0 & \cdots & 0 & -a_1 \\ -1 & t & 0 & \cdots & 0 & -a_2 \\ 0 & -1 & t & \cdots & 0 & -a_3 \\ 0 & 0 & -1 & \cdots & 0 & -a_4 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & t & -a_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & t-a_n \end{bmatrix} $
$+ (-1)^{n+1} \times (-a_0)(-1)^n $
$ = t[t^{n} - a_{n}t^{n-1} - a_{n-1}t^{n-2} - ... - a_3t^2 - a_2t - a_1] + (-1)^{2n+1} a_0$
$ = t^{n+1} - a_nt^n - a_{n-1}t^{n-1} - ... - a_2t^2 - a_1t - a_0$
Proof complete.
Best Answer
Expanding along the final column results in
$$a_n\cdot(-1)^{n+2} \begin{vmatrix} -y_1 & * & *&*&* \\ 0& -y_2& *&*&*\\ 0& 0 & -y_3&*&*\\ \vdots&\vdots&\vdots & \ddots&*\\ 0&0&0&\cdots &-y_n \end{vmatrix} + x_n\cdot \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_{n-1} \\ -y_1 & x_1 & 0 & \dots & 0 \\ 0 & -y_2 & x_2 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & x_{n-1} \end{vmatrix}$$
The first determinant is easy to calculate (it is $(-1)^n\cdot y_1\cdot y_2\cdots y_n$), while the second one is similar to the first, only smaller. So, if we define the determinant you wanted to calculate as $D_n$, you have
$$D_{n} = a_n\cdot y_1\cdot y_2\cdots y_n + x_nD_{n-1}$$
Now expanding that $D_{n-1}$ part further can yield some sort of solution (I don't see it being particularly pretty, however...)
Edit:
If I am not mistaken, the final result is
$$a_0x_1x_2\cdots x_n + a_1y_1x_2x_3\cdots x_n + \cdots + a_iy_1y_2\cdots y_i x_{i+1}\cdots x_n + \cdots + a_ny_1y_2\cdots y_n$$
or, written without all the dots:
$$\sum_{k=0}^n \left(a_k\prod_{i=1}^{k} y_i\prod_{i=k+1}^n x_i\right).$$
I don't see any obvious way to simplify this, however.