Let me add some detail to Pace's sketch.
Let's say that a pair $(A,B)$ of ideals of $R$ satisfies the conditions if
(i) $A$ and $B$ are proper,
(ii) $A+B=R$,
(iii) there is no idempotent $e$ in $A$ whose complementary idempotent
$1-e$ lies in $B$.
[Before beginning, note that Condition (i) follows from Condition (iii): using Condition (iii) and the fact that both $A$ and $B$ contain $0$, derive that neither $A$ nor $B$ can contain $1$.]
Lemma. If $(A,B)$ satisfies the conditions and $e\in A$ is idempotent, then $(A,B+Re)$ satisfies the conditions.
Pf. We establish Condition (iii) for $(A,B+Re)$ by reducing it to Condition (iii) for $(A,B)$. Assume that $f$ is an idempotent in $A$ whose complementary idempotent $1-f$ belongs to $B+Re$. Multiplying the expression ``$(1-f)\in B+Re$'' by $1-e$ yields that $(1-e)(1-f)\in B(1-e)+Re(1-e)=B(1-e)\subseteq B$, so the idempotent $(1-e)(1-f)$ belongs to $B$. But now, since $e, f\in A$, we have that the idempotent $1-(1-e)(1-f)=e+f-ef\in A$, while its complement lies in $B$. This contradicts Condition (iii) for $(A,B)$.
Condition (ii) for $(A,B+Re)$ holds because $B+Re$ extends $B$. \\\
Lemma. If $A$ and $B$ contain the same idempotents, $(A,B)$ satisfies the conditions, and $e$ is an idempotent with $e, 1-e\notin A$, then $(A+Re,B+Re)$ satisfies the conditions.
Pf. Let me argue only Condition (iii), since (i) is a consequence of (iii) and (ii) is obvious here.
Assume that there is some idempotent $f\in A+Re$ whose complementary idempotent $1-f\in B+Re$. Again we multiply throughout by $1-e$. We get $(1-e)f\in A(1-e)+Re(1-e)=A(1-e)\subseteq A$, and $(1-e)(1-f)\in B(1-e)\subseteq B$. Now, since we have assumed that $A$ and $B$ contain the same idempotents, and the idempotent $(1-e)f$ belongs to $A$, we must have $(1-e)f\in B$. Since both $(1-e)f, (1-e)(1-f)$ belong to $B$, the sum $(1-e)f+(1-e)(1-f)=1-e$ belongs to $B$, contrary to the assumptions of the lemma ($A$ and $B$ contain the same idempotents and $1-e\notin A$). \\\