In a preadditive category, an object is initial exactly when it is terminal

Assume $Z$ is initial. Then $\mathcal{C}(Z,Z)$ has exactly one element by [L2], so its identity morphism and its additive zero coincide. Hence $1_Z=0_{Z,Z}$.

Let $f:A\to Z$. Using step 1.1, $f=1_Z\circ f=0_{Z,Z}\circ f$. By bilinearity from [L1], the right-hand side is the zero element of $\mathcal{C}(A,Z)$. Thus every morphism $A\to Z$ equals the same zero arrow, so there is exactly one such morphism and $Z$ is terminal.

The same argument with arrows reversed shows that a terminal object is initial: if $Z$ is terminal then $\mathcal{C}(Z,Z)$ again has one element, so $1_Z=0_{Z,Z}$, and for any $g:Z\to A$ one has $g=g\circ 1_Z=g\circ 0_{Z,Z}=0$. Hence there is exactly one map from $Z$ to $A$.

Steps 2.1 and 2.2 prove the two implications, so initial and terminal are equivalent in a preadditive category.