Let $A=(a_{ij})_{1 \leq i,j \leq n}$ be an symmetric matrix with positive entries and $0$ on the diagonal, that is, $$ A = \begin{bmatrix} 0 & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{12} & 0 & a_{23} & \cdots & a_{2n} \\ a_{13} & a_{23} & 0 & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots \\ a_{1n} & a_{2n} & a_{3n} & & 0 \\ \end{bmatrix}. $$
We define the diagonal matrix $D$ whose entries are $d_i = \sum_{j = 1}^n a_{ij}$ and consider the matrix $$ B = \alpha \mathrm{Id} + D - A$$ for a real $\alpha > 0$.
$B$ is a matrix with dominant diagonal, and is therefore invertible. I need to show that the inverse $C = (c_{ij})$ of $A$ is such that $$c_{ii} \geq c_{ij}$$ for any $1 \leq i,j \leq n$.
The result is easily shown to be true for matrices of size $n=2$ for which $$A = \begin{bmatrix} 0 & a \\ a & 0 \\ \end{bmatrix}, \quad B = \begin{bmatrix} a + \alpha & - a \\ - a & a + \alpha \\ \end{bmatrix}, \quad C = \frac{1}{(\alpha + a)^2 - a^2} \begin{bmatrix} a + \alpha & a \\ a & a + \alpha \\ \end{bmatrix}.$$
The case $n=3$ is also simple but I have troubles showing the general case.
(My first attempt was to show the results on the cofactors of $A$, since the comatrix is positively proportional to the inverse, but I could not find a way to simply compare the cofactors.)
The result is true. In fact, it holds whenever $M$ is a strictly diagonally dominant symmetric M-matrix (or equivalently, when $M$ is a strictly diagonally dominant symmetric $Z$-matrix with a positive diagonal). In your symbols, this means the result holds when $M$ is in the form of $D'+D-A$, where $D$ and $A$ are defined in your way, except that the off-diagonal elements of $A$ are only required to be non-positive (rather than negative) and $D'$ is some positive diagonal matrix (that is not necessarily a positive multiple of $I$). Partition $M$ as $$ \pmatrix{m_{11}&-v^T\\ -v&S}. $$ Since $M$ is a strictly diagonally dominant M-matrix, we have $Se-v>0$. However, $S$ is also a strictly diagonally dominant M-matrix. Therefore $S^{-1}$ exists and it is entrywise non-negative. As the dot product of a non-negative and non-zero vector and a positive vector is positive, we have $S^{-1}(Se-v)>0$, i.e., $S^{-1}v<e$. Now, using Schur complement, we obtain $$ C:=M^{-1}=\pmatrix{(m_{11}-v^TS^{-1}v)^{-1}&\ast\\ (m_{11}-v^TS^{-1}v)^{-1}S^{-1}v&\ast}. $$ As $M$ is a strictly diagonally dominant M-matrix, it is positive definite. Therefore $c_{11}=(m_{11}-v^TS^{-1}v)^{-1}>0$. As every element of $S^{-1}v$ is smaller than $1$, we conclude that $c_{11}>c_{i1}$ for all $i\ne 1$.