This is probably not really what you are looking for, but it's the best I know. It's a strengthening of the fairly well-known Parikh's Theorem.
There is a characterisation of the bounded context-free languages. A language $L$ is bounded if $L\subseteq w_1^* w_2^* \ldots w_n^*$ for some fixed words $w_1,\ldots,w_n$, in which case we can define a corresponding subset of $\mathbb{N}_0^n$:
$\Phi(L) = \{(m_1,m_2,\ldots,m_n) \mid w_1^{m_1} w_2^{m_2} \ldots w_n^{m_n}\in L\}.$
By a theorem of Ginsburg and Spanier (Thm 5.4.2 in Ginsburg's book 'The Mathematical Theory of Context-free Languages'), a bounded language $L$ is context-free if and only if $\Phi(L)$ can be expressed as a finite union of linear sets, each with a stratified set of periods. For the definitions of the terms in the last sentence, see my MO question.
This characterisation can be very useful for proving (not necessarily bounded) languages not to be context-free. If we can find words $w_1,\ldots,w_n$ such that $L\cap w_1^*\ldots w_n^*$ is not context-free, then $L$ is not context-free either, since $w_1^*\ldots w_n^*$ is a regular language, and the intersection of a context-free language with a regular language is context-free.
HINT: I use the notation and terminology of the Wikipedia article on the Myhill-Nerode theorem. Show that for this language $L$, the equivalence relation $R_L$ has infinitely many equivalence classes. Use the fact that if $u,v\in L$, $u\ne v$, and $|u|=|v|$, then $uu\in L$, but $vu\notin L$, so $u$ is a distinguishing extension for $u$ and $v$.
Best Answer
Myhill-Nerode tells us, that a language $L$ is regular iff its Myhill-Nerode-Relation $\equiv_L$ given by
\[ x \equiv_L y :\!\iff \forall z \in \Sigma^*: xz \in L \leftrightarrow yz \in L \] has finitely many equivalence classes.
So to prove that our language $\mathrm{ADD}$ isn't regular we have to give an infinite set of pairwise inequivalent words. For $n \in \mathbb N$ let $x_n = 10^n \in \Sigma^*$. Then $\{x_n \mid n \in \mathbb N\}$ is an infinite set of words, we will now show that they are pairwise $\equiv_{\mathrm{ADD}}$-inequivalent. So let $n \ne m$, for $z_n = \mathord{+}0\mathord=10^n$ we have $x_nz_n = 10^n\mathord+0\mathord=10^n \in \mathrm{ADD}$, but $x_mz_n = 10^m\mathord+0\mathord=10^n \not\in \mathrm{ADD}$, so $x_n \not\equiv_{\mathrm{ADD}} x_m$ and hence $\mathrm{ADD}$ isn't regular.