Taking Jennifer Dylan's reasoning a tiny bit further:
Assume that $A$ is invertible, i.e. there are no zero eigenvalues. Then $A$ is similar to $\operatorname{diag}(J_{\lambda_1}(k_1),\dots,J_{\lambda_r}(k_r))$ where each $J_{\lambda_i}(k_i)$ is a Jordan block of size $k_i \times k_i$ and $\lambda_i \ne 0$. For any $\varepsilon>0$, the block $J_{\lambda_i}(k_i) = \begin{bmatrix} \lambda_i \\ 1 & \ddots \\ & \ddots & \ddots \\ & & 1 & \lambda_i \end{bmatrix}$ is similar to $\widetilde{J}_{\lambda_i}(k_i) = \begin{bmatrix} \lambda_i \\ \varepsilon & \ddots \\ & \ddots & \ddots \\ & & \varepsilon & \lambda_i \end{bmatrix},$ which you can easily verify by computing $P J_{\lambda_i}(k_i) P^{-1}$ where $P=\operatorname{diag}(1,\varepsilon,\dots,\varepsilon^{k_i-1}).$ For each $i$ simply pick $\varepsilon$ to be less than $|\lambda_i|$, and you've got a (strictly) diagonally dominant matrix. Note that if $A$ is singular then it cannot be similar to a strictly diagonally dominant matrix.
Remaining question: what do we do with Jordan blocks for $\lambda=0$?
In the $2 \times 2$ case we have $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}^{-1} = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix}. $