$$A=\begin{pmatrix}2&1&1\\0&1&0\\0&0&1\end{pmatrix}\;,\;\;A^{-1}=\begin{pmatrix}\frac12&\!\!-\frac12&\!\!-\frac12\\0&\;\;1&\;\;0\\0&\;\;0&\;\;1\end{pmatrix}$$
So $\,A\;$ is diag. dominant with all its entries non-negative, but its inverse is not even diag. dominant since in the first row
$$\left|\frac12\right|=\frac12\rlap{\,\;/}\ge \left|-\frac12\right|+\left|-\frac12\right|=1$$
Let $A$ be the matrix whose entries are $A_{i,j} = \begin{cases}a_{i} & \text{if} \ i = j \\ \alpha & \text{if} \ i \neq j\end{cases}$.
Let $D$ be the diagonal matrix with diagonal entries $D_{i,i} = a_i - \alpha$.
Then, $A = D + \alpha vv^T$ where $v$ is a $n \times 1$ vector of all ones.
If $A$ and $D$ are both invertible, then we may apply the Sherman-Morrison formula to get $$A^{-1} = (D+\alpha vv^T)^{-1} = D^{-1} - \dfrac{\alpha D^{-1} vv^T D^{-1}}{1 + \alpha v^TD^{-1}v}$$
Since $D$ is diagonal, the inverse of $D$ is simply a diagonal matrix with entries $(D^{-1})_{i,i} = (D_{i,i})^{-1} = \dfrac{1}{a_i - \alpha}$.
From here, it is easy to compute $A^{-1}$. The $i,j$-th entry of $A^{-1}$ is given by $$(A^{-1})_{i,j} = \begin{cases}\dfrac{1}{a_i-\alpha} - \dfrac{1}{c(a_i-\alpha)^2} & \text{if} \ i = j \\ - \dfrac{1}{c(a_i-\alpha)(a_j-\alpha)} & \text{if} \ i \neq j\end{cases},$$
where $c = \dfrac{1}{\alpha} + \displaystyle\sum_{i = 1}^{n}\dfrac{1}{a_i-\alpha}$.
Of course, this doesn't handle the case where $D$ isn't invertible, which occurs precisely when $a_i = \alpha$ for some $i$.
Best Answer
The Sherman-Morrison formula gives the inverse of any rank one update of a matrix for which the inverse is known.
Here we can write your matrix as follows:
$$ \begin{pmatrix} a & -b & -b \\ -b & c & -b \\ -b & -b & d \end{pmatrix} = \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix} - b \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} $$
Since the first summand is an invertible diagonal matrix (regardless of the actual sign of $b$), we have that the Sherman-Morrison formula can be applied.
Let $A = \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix}$ and let $u = \begin{pmatrix} -b \\ -b \\ -b \end{pmatrix}$, $v^T = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix}$. What you ask for is:
$$ (A + uv^T)^{-1} = A^{-1} - \left( \frac{1}{1+ v^T A^{-1} u} \right) \left( A^{-1} uv^T A^{-1} \right) $$
Note that the first factor in the second term of the right hand side is just a scalar, obtained by taking the reciprocal of the scalar $1+ v^T A^{-1} u$. It multiplies the rank one matrix $A^{-1} uv^T A^{-1}$, showing that the inverse of the rank one update of $A$ is itself a rank one update of $A^{-1}$.