Q: If, for elements $A$ and $B$ of a ring $R$, we have
(i) $AB=0$, and
(ii) $T:=A+B$ is invertible,
then how can we verify the existence of $C$ and $D$ in $R$ such that $CA+BD=1$?

Let me start by describing an explicit example of a ring $R$ having elements $A$ and $B$ satisfying (i) and (ii), yet the desired elements $C$ and $D$ do not exist. Then I will describe some extra conditions which guarantee the existence of $C$ and $D$.

Let $\Sigma=\mathbb Z^{\mathbb Z^+}$ be the abelian group whose elements are the sequences $\overline{m}=(m_1,m_2,m_3,\ldots)$ of integers indexed by the positive integers. Define endomorphisms $A$ and $B$ of this abelian group by:

$$A\overline{m} = (m_1,m_3,m_5,0,m_7,m_9,m_{11},0,m_{13},m_{15},m_{17},0,\ldots)$$

$$B\overline{m} = (0,0,0,m_2,0,0,0,m_4,0,0,0,m_6,0,0,0,m_8,0,\ldots).$$

That is, $A$ ignores the even coordinates of $\overline{m}$, and orders the odd coordinates in the expected order, except inserts a $0$ at the coordinates indexed by a multiple of $4$. $B$ ignores the odd coordinates, orders the even coordinates in the expected order, but inserts $0$'s everywhere EXCEPT the coordinates indexed by a multiple of $4$. The main points to note are that

($\alpha$) $\textrm{im}(B)$ is (properly) contained in $\textrm{ker}(A)$, so in particular $AB$ is the zero endomorphism, and
($\beta$) the endomorphism $T:=A+B$ acts on the group $\Sigma$ by permutation of coordinates.

Thus, if $R=\textrm{End}(\Sigma)$, then $R$ meets all of conditions of the question. But $R$ contains no elements $C$ and $D$ such that $CA+BD=1$. Such a relation would force $\textrm{ker}(A)\subseteq \textrm{im}(B)$, which is false. Specifically, $$e_2=(0,1,0,0,\ldots)\in \textrm{ker}(A)\setminus \textrm{im}(B),$$ and this is in conflict with $CA+BD=1$, since $(CA+BD)e_2 = BDe_2\in\textrm{im}(B)$, while $1e_2=e_2\notin \textrm{im}(B)$.

Now let me mention some finiteness conditions which can be added to the problem to guarantee the existence of $C$ and $D$ so that $CA+BD=1$.

(1) $T$ is not just invertible, but it is a unit of finite order. OR,
(2) $T^{-1}$ belongs to the subring of $R$ generated by $A$ and $B$. OR,
(3) $T^{-1}B\in BR$. OR,
(4) $BR=B^2R$.

Note that if (1) holds, and $T$ is a unit of order $n$, then $T^{-1}=T^{n-1}=(A+B)^{n-1}\in \langle A, B\rangle$ and (2) holds. Note that if (2) holds and if $R$ is generated by $A$ and $B$, then the right ideal $BR$ is actually a $2$-sided ideal, since it is closed under left multiplication by both $A$ and $B$. In fact, $BR$ is the $2$-sided ideal $(B)$ generated by $B$. Thus $T^{-1}B\in (B)=BR$, and (3) holds. Now, if (3) holds, we have $T^{-1}B=BD$ for some $D\in R$, so $B=TBD=(A+B)BD=B^2D\in B^2R$, which is enough to establish that $BR=B^2R$.

Thus condition (4) is essentially the most general of these. Let me add that condition to the conditions of original problem:

Claim. If, for elements $A$ and $B$ of a ring $R$, we have
(i) $AB=0$,
(ii) $T:=A+B$ is invertible, and
(ii) $BR=B^2R$,
then $R$ has elements $C$ and $D$ such that $CA+BD=1$

Reasoning. $A+B=T$, so $T^{-1}A+T^{-1}B=1$. We have added the assumption that $BR=B^2R$, so $B=B^2D$ for some $D\in R$. Thus $B=B^2D = (A+B)BD = TBD$, and we get $T^{-1}B=BD$. Substituting this into the equation from the first line of this argument we get $T^{-1}A+BD=1$. Now let $C=T^{-1}$ to convert this to $CA+BD=1$. \\\\\