Schur's Lemma in Group Theory

Nice question! The answer is no. The smallest potential counterexample works.

Let $G$ be the binary icosahedral group. This is a perfect group of order $120$ (in fact the smallest nontrivial such group which is not simple). Its only nontrivial normal subgroup is its center $\pm 1$, hence its only nontrivial quotient is the icosahedral group $A_5$, which cannot occur as a subgroup of $G$, since any subgroup of $G$ of order $60$ is necessarily normal. Hence $G$ is not simple but every nonzero homomorphism $G \to G$ is an isomorphism.

Dan Shved gives an example for your second question in the comments.

By request, posting a counterexample for question 2 as an answer.

The group of rational numbers $\mathbb{Q}$ under addition is clearly not simple. But the endomorphism ring $\mathrm{End}(\mathbb{Q})$ is isomorphic to $\mathbb{Q}$ itself, and is a division ring, which means that all non-zero endomorphisms are automorphisms.