I'm trying to show that assuming that $s_{1}\leq\tau$ and $s_{2}\leq\tau$ and all parameters are positive, \begin{eqnarray*} a_{1}b_{1} & \geq & a_{2}b_{2},\\ a_{1}(\tau+b_{1}\frac{\tau^{2}}{2}) & \geq & a_{2}(\tau+b_{2}\frac{\tau^{2}}{2}),\\ a_{1}(s_{1}+b_{1}\frac{s_{1}^{2}}{2}) & = & a_{2}(s_{2}+b_{2}\frac{s_{2}^{2}}{2}), \end{eqnarray*} implies that $$ a_{1}(\tau+b_{1}\frac{\tau^{2}}{2})+a_{1}b_{1}s_{1}\tau\geq a_{2}(\tau+b_{2}\frac{\tau^{2}}{2})+a_{2}b_{2}s_{2}\tau, $$ Mathematica returns an empty list when asked to find a counter example, but I haven't been able to show that the inequality has to hold.
Edit: Dividing all equations by $a_1$ and denoting $z=a_1/a_2$ the problem is showing that assuming $s_{1}\leq\tau$ and $s_{2}\leq\tau$, \begin{eqnarray*} b_{1} & \geq & zb_{2},\\ \tau+b_{1}\frac{\tau^{2}}{2} & \geq & z(\tau+b_{2}\frac{\tau^{2}}{2}),\\ s_{1}+b_{1}\frac{s_{1}^{2}}{2} & = & z(s_{2}+b_{2}\frac{s_{2}^{2}}{2}), \end{eqnarray*} implies that $$ \tau+b_{1}\frac{\tau^{2}}{2}+b_{1}s_{1}\tau\geq z(\tau+b_{2}\frac{\tau^{2}}{2}+b_{2}s_{2}\tau), $$
The following discusses a related problem where the conclusion is replaced with:
which reduces to the original problem for $\lambda = 1$.
The inequality is proved to hold for $\lambda = \frac{1}{2}$. Also, a sufficient condition is derived for it to hold for $\lambda = 1$, however that condition is not obviously implied by the given premises alone.
With the substitutions for $k=1,2$:
$$u_k = a_k \;\tau\;,\quad v_k=\frac{1}{2}a_k\;b_k\;\tau^2\;,\quad t_k = \frac{1}{\tau}s_k \;\le\; 1$$
the problem can be restated as: given that
$$v_1 \ge v_2 \tag{1}$$
$$u_1+v_1 \ge u_2+v_2 \tag{2}$$
$$u_1\,t_1 + v_1\,t_1^2 = u_2\,t_2 + v_2\,t_2^2 \tag{3}$$
prove that
$$u_1 + v_1 + 2 \lambda v_1 t_1 \ge u_2 + v_2 + 2 \lambda v_2 t_2 \tag{4}$$
Case (a): assume $u_1 \lt u_2$. Then note that if both $t_1 \lt t_2$ and $v_1 t_1 \lt v_2 t_2$ hold, then $u_1\,t_1 + v_1\,t_1^2 = t_1(u_1 + v_1t_1)\lt t_2(u_2 + v_2t_2) = u_2\,t_2 + v_2\,t_2^2$ because each term on the LHS is smaller than the corresponding one on the RHS. But that would contradict $(3)$, so at least one of $t_1 \ge t_2$ or $v_1 t_1 \ge v_2 t_2$ must hold true. The former implies the latter given $(1)$, so $v_1 t_1 \ge v_2 t_2$ in either case, and $(4)$ follows by adding $(2)$ and $2 \lambda v_1t_1 \ge 2 \lambda v_2t_2$. This proves case (a) for $\forall \lambda \gt 0$.
Case (b): assume $u_1 \ge u_2$. Given $(1)$, for $(3)$ to hold it must be that $t_1 \le t_2$ (otherwise the LHS would be strictly greater than the RHS). It follows that:
$$u_1 + v_1t_1 = \frac{t_2}{t_1}(u_2 + v_2t_2) \ge u_2 + v_2t_2 \tag{5}$$
Rewriting $(4)$ as:
$$(1 - 2 \lambda)u_1 + v_1 + 2 \lambda (u_1 + v_1 t_1) \ge (1 - 2 \lambda)u_2 + v_2 + 2 \lambda (u_2 + v_2 t_2)$$
and given $(5)$ it follows that a sufficient condition for the inequality to hold is:
$$(1 - 2 \lambda)u_1 + v_1 \ge (1 - 2 \lambda)u_2 + v_2 \tag{6}$$
For $\lambda = \frac{1}{2}\;$, $(6)$ reduces to $(1)$ and is always satisfied, so the inequality $(4)$ is true for $\lambda = \frac{1}{2}$.
For $\lambda = 1\;$ (OP's original question) a sufficient condition $(6)$ for the inequality to hold is $v_1 - u_1 \ge v_2 - u_2$ which, however, does not directly follow from the given premises.