= Solution
The subspace spanned by $e_4,e_5,e_6$ is the <center of a Lie algebra>. Each of their tensor squares is therefore invariant. On tensor products the infinitesimal <Adjoint representation of a Lie algebra> acts by a derivation, $D_x(u\otimes v)=[x,u]\otimes v+u\otimes[x,v]$. Write $u\odot v=u\otimes v+v\otimes u$ and let $S=e_1\odot e_5+e_3\odot e_6-e_2\odot e_4$. The only potentially nonzero checks are
$$
D_{e_1}S=e_4\odot e_6-e_6\odot e_4=0,\quad
D_{e_2}S=-e_6\odot e_5+e_5\odot e_6=0,\quad
D_{e_3}S=-e_4\odot e_5+e_5\odot e_4=0.
$$
The remaining generators are central, so they act trivially. This proves every claimed tensor invariant.
These tensors are contravariant. To obtain an ordinary <metric tensor>, take an invertible linear combination and invert it. For arbitrary real $a_4,a_5,a_6$ and $t\ne0$, set
$$
Q=tS+a_4e_4\otimes e_4+a_5e_5\otimes e_5+a_6e_6\otimes e_6.
$$
In the ordered basis $(e_1,e_2,e_3;e_4,e_5,e_6)$ its matrix is
$$
Q=\begin{pmatrix}0&tK\\tK^T&A\end{pmatrix},\qquad
K=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix},\qquad A=\operatorname{diag}(a_4,a_5,a_6).
$$
Since $K^TK=I$, $\det Q=-t^6\ne0$, and
$$
Q^{-1}=\begin{pmatrix}-t^{-2}KAK^T&t^{-1}K\\t^{-1}K^T&0\end{pmatrix}.
$$
Let $\lambda^a$ be the dual left-invariant coframe. The resulting <neutral invariant metrics on the free two-step Lie algebra on three generators> are
$$
\boxed{g=t^{-1}(\lambda^1\odot\lambda^5+\lambda^3\odot\lambda^6-\lambda^2\odot\lambda^4)
-t^{-2}\bigl[a_5(\lambda^1)^2+a_4(\lambda^2)^2+a_6(\lambda^3)^2\bigr].}
$$
Here $(\lambda^a)^2=\lambda^a\otimes\lambda^a$. Infinitesimal invariance of $Q$ implies that its inverse is an adjoint-invariant <bilinear form>. Its left-translated metric is right-invariant on a connected $G$, giving the desired \b[four-parameter family of bi-invariant metrics]. A change of variables removes the central $A$ block from the quadratic form, leaving three off-diagonal two-dimensional blocks. Each has one positive and one negative eigenvalue. Thus the <metric signature> is $(3,3)$; these are <bi-invariant pseudo-Riemannian metrics>, not positive-definite ones.
Connectedness, conventionally understood here, matters for the passage from infinitesimal to full group invariance. If $G$ is disconnected, its other components must preserve $Q$ separately. For example the <Lie algebra automorphism> reversing $e_1,e_4,e_6$ and fixing $e_2,e_3,e_5$ preserves all brackets but sends $S$ to $-S$. In the semidirect product of the simply connected group with this order-two automorphism, none of these nondegenerate $t\ne0$ tensors is invariant under that component. Thus the Lie-algebra data alone give the stated family on the identity component, rather than an unconditional extension to every disconnected realization.
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