This is not as simple as Travis Willse's approach. It's a bit more hands on, which means it takes a bit more work, but it may give some insight into why the left-invariant one-forms appear as they do.
Let $M \in G$ be a point with coordinates $(a, b, c_1, c_2, d_1, d_2)$. Then there is a diffeomorphism $f_M : G \to G$ given by $A \mapsto MA$. A one form $\eta$ on $G$ is left-invariant if $f_M^*\eta = \eta$ for all $M \in G$. Performing the matrix multiplication $MA$, one see that in the given coordinates, we have
$$f_M(t, x, y_1, y_2, z_1, z_2) = (t + a, x + b, y_1e^a + z_1be^a + c_1, y_2e^{-a}+z_2be^{-a} + c_2, z_1e^{a} + d_1, z_2e^{-a} + d_2).$$
Therefore
\begin{align*}
f_M^*dt &= d(t+a) = dt\\
f_M^*dx &= d(x+b) = dx\\
f_M^*dy_1 &= d(y_1e^a + z_1be^a + c_1) = e^ady_1 + be^adz_1\\
f_M^*dy_2 &= d(y_2e^{-a}+z_2be^{-a} + c_2) = e^{-a}dy_2 + be^{-a}dz_2\\
f_M^*dz_1 &= d(z_1e^{a} + d_1) = e^adz_1\\
f_M^*dz_2 &= d(z_2e^{-a} + d_2) = e^{-a}dz_2.
\end{align*}
So we see that $\alpha := dt$ and $\beta := dx$ are left-invariant.
The last two equations have a right hand side which is very similar to the original form, so let's focus on those next. The fact that the right hand side is a rescaling of the original form indicates that $dz_i$ are not left-invariant, but we can make them left-invariant by multiplication with an appropriate function. As the coefficient of the right hand side only depends on $a$, we're looking for functions $g_i(t)$ such that $g_i(t)dz_i$ are left-invariant. Note that $f_M^*(g_2(t)dz_2) = g_2(t+a)e^{-a}dz_2$. To obtain a left-invariant form, we need $g_2(t+a)e^{-a} = g_2(t)$, and hence $g_2(t+a) = g_2(t)e^a$, so we must have $g_2(t) = e^t$. Likewise, $g_1(t)dz_2$ is left-invariant if and only if $g_1(t+a) = g_1(t)e^{-a}$, so $g_1(t) = e^{-t}$. Therefore $\delta_1 := e^{-t}dz_1$ and $\delta_2 := e^t dz_2$ are left-invariant.
Finally, let's consider the middle two equations. By the considerations in the previous paragraph, we can rescale by $e^{\mp t}$ to deal with the common $e^{\pm a}$ factor:
\begin{align*}
f_M^*(e^{-t}dy_1) &= e^{-t}dy_1 + be^{-t}dz_1 = e^{-t}dy_1 + b\delta_1\\
f_M^*(e^tdy_2) &= e^tdy_2 + be^tdz_2 = e^tdy_2 + b\delta_2.
\end{align*}
The only way that $b\delta_i$ can arise from the pullback of a form is if that form involves a term $h_i(x)\delta_i$ (compare with the discussion in the previous paragraph). As
$$f_M^*(e^{-t}dy_1 + h_1(x)\delta_1) = e^{-t}dy_1 + b\delta_1 + h_1(x+b)\delta_1 = e^{-t}dy_1 + (h_1(x+b) + b)\delta_1,$$
the form $e^{-t}dy_1 + h_1(x)\delta_1$ is left-invariant if and only if $h_1(x+b) = h_1(x) - b$, so we must have $h_1(x) = -x$. Likewise, $e^tdy_2 + h_2(x)\delta_2$ is left-invariant if and only if $h_2(x+b) = h_2(x) - b$ so $h_2(x) = -x$. Therefore $\gamma_1 := e^{-t}dy_1 - x\delta_1 = e^{-t}dy_1 -xe^{-t}dz_1$ and $\gamma_2 := e^tdy_2 - x\delta_2 = e^tdy_2 - xe^tdz_1$ are left-invariant.
