Sylvester's criterion says that an $n\times n$ Hermitian matrix $A$ is positive definite if and only if all its leading principal minors of are positive. If one knows that fact that every Hermitian matrix has an orthogonal eigenbasis, one can prove Sylvester's criterion easily.
The forward implication is obvious. If $A$ is positive definite, so are all its leading principal submatrices. Their spectra are hence positive. Thus the leading principal minors are positive, because each of them is a product of the eigenvalues of the submatrix.
To prove the backward implication, we use mathematical induction on $n$. The base case $n=1$ is trivial. Suppose $n\ge2$ and all leading principal minors of $A$ are positive. In particular, $\det(A)>0$. If $A$ is not positive definite, it must possess at least two negative eigenvalues. As $A$ is Hermitian, there exist two mutually orthogonal eigenvectors $x$ and $y$ corresponding to two of these negative eigenvalues. Let $u=\alpha x+\beta y\ne0$ be a linear combination of $x$ and $y$ such that the last entry of $u$ is zero. Then
$$
u^\ast Au=|\alpha|^2x^\ast Ax+|\beta|^2y^\ast Ay<0.
$$
Hence the leading $(n-1)\times(n-1)$ principal submatrix of $A$ is not positive definite. By induction assumption, this is impossible. Hence $A$ must be positive definite.
We will make essential use of the matrix determinant lemma. The key result is the following proposition. I've included a full proof, so try not to read past the initial proposition if you want to try your hand at it yourself.
Proposition: Let $A$ be an $n\times n$ matrix such that all principal minors are positive. Then the matrix
$$A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k$$
has positive determinant for all $\lambda \ge 0$.
Here $\mathbf{e}_k\in \mathbb{R}^n$ is the $k$th standard basis vector of $\mathbb{R}^n$.
Proof: This is a direct application of the matrix determinant lemma, which says that we have
$$\det(A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k) = \det(A) + \lambda[\mathrm{Adj}(A)]_{kk},$$
where $[\mathrm{Adj}(A)]_{kk}$ denotes the $kk$th entry of the adjugate matrix of $A$. Explicitly, we know that
$$[\mathrm{Adj}(A)]_{kk} = C_{kk},$$
where $C_{kk}$ is the $kk$th cofactor of $A$. But this is a principal minor of $A$, which by assumption was positive. Therefore it follows that we have
$$\det(A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k) = \det(A) + \lambda[\mathrm{Adj}(A)]_{kk} \ge \det(A) > 0,$$
where the $\geq$ sign becomes an equality if and only if $\lambda = 0$. $\square$
Corollary: Let $A$ be an $n\times n$ matrix such that all principal minors are positive. Then the matrix
$$B=A + \lambda\mathbf{e}_k\mathbf{e}^\mathrm{T}_k$$
also has all principal minors positive for $\lambda \ge 0$.
Proof: Consider a principal minor $[B]_I$, where $I$ is the index set of the rows/columns retained when forming the minor. If $k\notin I$, then the principal minor has not changed, i.e. we have $[B]_I = [A]_I > 0$. If $k\in I$, then apply the proposition to the underlying submatrix. $\square$
Finally, we can prove your result.
Proposition: Let $A$ be an $n\times n$ matrix with all positive principal minors. Let $D$ be a non-negative diagonal matrix with diagonal entries $d_k$. Then $A+D$ has all principal minors positive.
Proof: We can simply update $A$ one step at a time using the corollary. First, since $A$ has all principal minors positive, it follows that
$$A_1 = A + d_1\mathbf{e}_1\mathbf{e}_1^\mathrm{T}$$
also has all principal minors positive. Now apply the corollary again to
$$A_2 = A_1 + d_2\mathbf{e}_2\mathbf{e}_2^\mathrm{T},$$
and inductively, to
$$A_k = A_{k-1} + d_k\mathbf{e}_k\mathbf{e}_k^\mathrm{T},$$
and conclude that each $A_k$ has all positive principal minors. The desired result follows by noting that $A+D = A_n$. $\square$
Best Answer
What you are suggesting is that the (necessary and sufficient) conditions
for an $n\times n$ matrix $A$ to be an M-matrix are equivalent to (or implied by) the conditions
Note that the positivity of $1\times 1$ minors means that $A$ has positive diagonal.
This is true trivially if $n=1$ or $n=2$. It is also true if $n=3$ (positivity of diagonal and non-negativity of off-diagonal entries implies positivity of $2\times 2$ minors). However, this fail to be the case when $n\geq 4$. For example, $$ A=\left(\begin{array}{rrrr} 1 & -2 & -1 & -3\\ -1 & 2 & -5 & -2\\ -1 & -5 & 1 & -1\\ -5 & -1 & -1 & 3 \end{array}\right). $$ Note that the determinants of both the leading and trailing $3\times 3$ submatrices are negative but $\det(A)=30>0$.