A Lie group has three standard Cartan connections; the ()connection, the (0)connection, and the (+)connection. The (0)connection is LeviCivita with the associated metric the biinvariant metric. The other two connections aren't LeviCivita due to the presence of torsion. However, there's nothing to stop them a priori from being metric connections. My question is; are the minus and plus connections compatible with the biinvariant metric? This seems reasonable but I can't find a reference.

Yes. Let $\nabla$ be an arbitrary connection on the tangent bundle of a Riemannian manifold $(M,g)$. The standard trick for expressing the LeviCivita connection in terms of $g$ gives you, for any 3 vector fields $X$, $Y$, $Z$: $$Xg(Y,Z)+ Yg(Z,X) Zg(X,Y)= N(X,Y,Z) $$ $$+ g(T(X,Z),Y)+ g(T(Y,Z),X) g(T(X,Y),Z) $$ $$ +2 g(\nabla_X Y,Z) g([X,Y],Z) + g([X,Z],Y) + g([Y,Z],X),$$ where $$ T(X,Y)=\nabla_X Y \nabla_Y X [X,Y]$$ is the torsion of $\nabla$ and $$ N(X,Y,Z)= \nabla_Xg(Y,Z)+ \nabla_Yg(Z,X)\nabla_Zg(X,Y). $$ This is the "nonmetricity": $N=0\Leftrightarrow \nabla g=0$. Now, turning to the case at hand: we define the $\pm$ and $0$ connections by $$ (\nabla_X Y)_e=\epsilon [X,Y],$$ $ \epsilon = 1, 0, \frac{1}{2}$ respectively, so the torsion is $$T(X,Y) = (2\epsilon 1)[X,Y]= \pm[X,Y]\textrm{ or } 0, $$ hence the names of the connections. But then you get $$ 0 = N(X,Y,Z) 2\epsilon\left[ g([Z,Y],X) + g(Y,[Z,X]) \right],$$ and the second summand is zero due to biinvariance, so $N=0$. 


If I understand correctly, the answer is Yes. the +//0 connections can be defined by, if $X,Y$ is the left invariant vector $$\nabla_{X}Y=a[X,Y]$$ where $a=1,1,0$. The connection is metric for left invariant metric iff $$0=\langle\nabla_{X}Y,Z\rangle+\langle Y,\nabla_{X}Z\rangle.$$ This is trival for the biinvariant metric. 

