Lie algebras role in algebraic group theory I know nothing about algebraic groups, but I know somewhat about Lie groups. I'm wondering what their exact analogue is in algebraic groups.
Given $G$, we still have the tangent space at $e$, call it $T$. We have $Ad:G \to GL(T)$, and so we have a map $T \to End(T)$, so lie bracket is still defined.
Suppose $G, H$ are algebraic groups (connectedish?) is a map $G \to H$ determined by the lie algebra map? Is there a baker formula?
The reason I am concerned this might not be true is I saw a proof in math overflow that a projective  irreducible algebraic group is abelian; the proof noted that by properness, $Ad$ is trivial (Being the image of a map from proper to affine), but while in regular Lie groups this would suffice to finish, he needed to note that it must also fix either order derivatives; (i.e the stalk mod a higher power of the maximal ideal).
 A: I mean this is a pretty vague question. Lie algebras play a pivotal role in the theory of algebraic groups. Specifically, the classification of simply connected semisimple groups over an algebraically closed field is still in terms of root data which, as you might expect from Lie group theory, are defined in terms of the adjoint action of a maximal torus. Even arithmetically root data play an important role in the form of relative root data which can help measure how 'non-split' a group is, as well as classifying most of the important data of the group. One can look at Brian Conrad's notes on the topic (but beware that they are quite advanced).
Let me try and answer your specific questions.

Fact: Let $f,g:G\to H$ be smooth morphisms of algebraic groups with $G$ connected. Suppose that $df=dg$, then $f=g$. If $g=\mathrm{id}$ then one only requires that $\ker(f)$ is smooth (over $k$).

Proof: By considering $fg^{-1}$ it suffices to prove the second claim. But, by considering $\ker(f)$ it suffices to show that if $K\subseteq G$ is a smooth subgroup such that $\mathrm{Lie}(K)=\mathrm{Lie}(G)$, then $K=G$. This follows from [Mil, Proposition 10.15]. $\blacksquare$
This answer your first question I believe.

Is there a Baker(--Campbell--Hausdorff) formula?

Well, since our groups are algebraic the answer is no--infinite series don't make sense. But, infinite series do make sense when one plugs nilpotent things into them. In particular, for unipotent algebraic groups (which have nilpotent Lie algebras) one gets an algebraic formula from Baker--Campbell--Hausdorff power series (see [Mil, 14.35])--at least in characteristic $0$ so that we can have the correct denominators. This actually allows one to establish an equivalence (in characteristic $0$) between unipotent algebraic groups and nilpotent Lie algebras (see [Mil, Theorem 14.37]).

Why can one not apply the above fact to prove abelian varieties are commutative?

I assume that the proof you saw is similar to the proof given in Theorem 1.5.1 here. The reason that one can not apply the above fact is that in fully generality you don't know that the kernel of $x\mapsto axa^{-1}=C_G(a)$ is a smooth subgroup of $A$. In characteristic $0$ this is guaranteed (by [Mil, Theorem 3.23]), but since loc. cit. is interested in general characteristic such smoothness cannot be guaranteed.
References:
[Mil] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
