Constructing a metric on a Lie Group
There is such a (bi-invariant) metric on a Lie group (isomorphic to a compact Lie group times $\mathbb{R}^{n}$). It is the one induced by the Killing form. For more input on this, see the mathoverflow discussion on the topic:
https://mathoverflow.net/questions/32554/why-the-killing-form
It should be pointed out that you do not need the exponential map to construct the metric, which is defined on the tangent space $T_{p}g, g\in G$, not on the group itself. All you need is a nicely behaved symmetric bilinear form under the group action. The explicit construction can be found at here.
Update: Yor pointed out in the comments that for a non-semisimple compact Lie group there is a bi-invariant metric, but it is not induced by the Killing form, as the Killing form degenerates on the center. To solve this, write $\mathfrak{g}=\mathfrak{a}\times \mathfrak{s}$. The Killing form induces a metric on $\mathfrak{s}$ and we put an arbitrary metric on $\mathfrak{a}$.
You ask about the metric tensor of a Lie group, but such a thing does not exist. What you can do, though, is to (arbitrarily) choose a scalar product $\langle, \rangle _e$ in $T_e G$ and then by (left) translations define $\langle, \rangle _g = (L_g) _* \langle, \rangle _e$, that is for $u,v \in T_g G$ define $\langle u, v \rangle _g = \langle \mathrm d L_g ^{-1} (u), \mathrm d L_g ^{-1} (v) \rangle _e$, where $L_g (x) = gx$. This is a left-invariant Riemann tensor, which in turn gives rise to a left-invariant distance $d$, i.e. $d(gx, gy) = d(g,h)$ for all $g,x,y \in G$. Neither the metric tensor, nor the associated distance need be right-invariant (and, symetrically, if you construct them to be right-invariant, they need not be left-invariant).
The following result is taken from Chapter 17 of "Linear Algebra, Manifolds, Riemannian Metrics, Lie Groups and Lie algebra, with Applications to Robotics, Vision and Machine Learning" by Jean Gallier:
If $G$ is connected, $\langle, \rangle$ defined above is bi-invariant (meaning left- and right-invariant) if and only if $\langle [u,v], w \rangle = - \langle v, [u,w] \rangle$ for all $u,v,w \in T_e G$.
If $G$ is compact, then one may define a bi-invariant Riemann tensor $\langle, \rangle '$ from a left- (or right-) invariant one $\langle, \rangle$ by Haar-integrating the latter: $\langle u, v \rangle ' = \int _G \langle \mathrm d L_g u, \mathrm d L_g v \rangle_{gx} \ \mathrm d g$ for all $u,v \in T_x G$ and $x \in G$.
If $G$ is semisimple, then its Lie algebra $\mathfrak g \simeq T_e G$ possesses a preferred scalar product - the Kiling form (or its opposite, depending on the sign convention used), and this in turn gives rise to a preferred metric tensor on $G$.