I came up with an argument that seems to prove a result I can't find anywhere. Did I make a mistake somewhere or is this right?
Suppose that $F:\mathcal{A}\to \mathcal{B}$ is a fully faithful functor between abelian categories that preserves zero objects. I claim that it must be additive, i.e. it preserves direct sum. Suppose that $\iota_1,\iota_2,\pi_1,\pi_2$ are the inclusion and projection maps between $A_1,A_2$ and their direct sum $A_1\oplus A_2$, so $\pi_1\circ \iota_1=1_{A_1},\pi_2\circ \iota_2=1_{A_2},\pi_1\circ \iota_2=0,\pi_2\circ \iota_1=0,$ and $\iota_1\circ \pi_1+\iota_2\circ \pi_2=1_{A_1\oplus A_2}$. Because $F$ preserves zero morphisms, it will preserve the first four identities. To prove that it preserves the final one, let $y$ be such that $F(y)=F(\iota_1)\circ F(\pi_1) +F(\iota_2)\circ F(\pi_2)$. Then $$F(\pi_1\circ y) =F(\pi_1)\circ (F(\iota_1)\circ F(\pi_1) +F(\iota_2)\circ F(\pi_2)) =F(\pi_1).$$ By faithfulness $\pi_1\circ y=\pi_1$. Similarly $\pi_2\circ y=\pi_2$. Because $\pi_1,\pi_2$ are projections from the direct sum $A_1\oplus A_2$, $y=1_{A\oplus B}$ so $F(\iota_1)\circ F(\pi_1) +F(\iota_2)\circ F(\pi_2)=F(y)=1_{F_{A_1\oplus A_2}}.$ Then $F$ preserves all five identities and so takes direct sum to direct sum. Am I missing something?