Infinite Krull-Schmidt categories?

The statement about simple Lie algebras is not true.

A (finitely generated right) module $P$ for a ring $R$ is stably free if $P\oplus R^m\cong R^n$ for some integers $m,n$.

Suppose $R$ has a non-free stably free module $P$, as above. If also $R$ is a (right) Noetherian domain, then the regular module $R$ is indecomposable and $P$ is a finite direct sum of indecomposable modules. So $R^n$ has two distinct decompositions into finitely many indecomposable summands, contradicting the Krull-Schmidt property.

If $\mathfrak{g}$ is any finite dimensional Lie algebra, then its universal enveloping algebra $U(\mathfrak{g})$ is a Noetherian domain.

Probably there are particular examples that predate this, but Theorem 2.6 of

Stafford, J. T., Stably free, projective right ideals, Compos. Math. 54, 63-78 (1985). ZBL0565.16012

shows that if $\mathfrak{g}$ is a finite dimensional non-abelian Lie algebra, then $U(\mathfrak{g})$ always has a non-free (finitely generated) stably free module.

In fact, I think that the construction gives an indecomposable non-free $P$ with $P\oplus U(\mathfrak{g})\cong U(\mathfrak{g})\oplus U(\mathfrak{g})$.

Theorem 1 of Chapter 1 of Pierre Gabriel's famous paper [Des catégories abéliennes, Bull. Soc. Math. France 90 (1962), 323–448] says that one gets a nice Krull-Schmidt theorem for arbitrary direct sums of objects in an abelian category $\mathcal A$ if $\mathcal A$ has a set of generators and exact "inductive" limits, and we are considering direct sums of indecomposable objects having local endomorphism rings. I am guessing that this applies to your situation. (I don't know much about the endomorphism rings of infinite dimensional indecomposables in your category.)