By [Pólya’s theorem][1], any even real-valued function $f$ on $\mathbb R$ with $f(\infty-)=0$ which is convex on $[0,\infty)$ is positive definite. So, any such function is the (auto)covariance function of a stationary Gaussian process; see e.g. [Section "Properties of the Autocovariance Function", page 2][2]. 

Now just take any two different functions, $f_1$ and $f_2$, of the Pólya class such that $f_2(t)=1-|t|=f_2(t)$ for $|t|\le1/2$. Then the corresponding stationary Gaussian processes, say $(X_{1,t})$ and $(X_{2,t})$, with the covariance functions $f_1$ and $f_2$ will have different distributions. Therefore, these two processes will be different from each other. 

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To be more specific, note first here that, 
by vertical and horizontal re-scaling, without loss of generality $a_1=a_2=1$, so that
$$EX_sX_t=1-|t-s|\quad\text{if}\quad|t-s|\le u, \tag{1}$$
where $u\in(0,1)$. 
Let then 
$$Y_t:=B_{t+1}-B_t=\int_t^{t+1}dB_s,$$
where $(B_t)_{t\in\mathbb R}$ is the standard Brownian motion with $B_0=0$. Then 
$$EY_sY_t=1-|t-s|\quad\text{if}\quad|t-s|\le 1,$$
so that 
$$EY_sY_t=EX_sX_t\quad\text{if}\quad|t-s|\le u,$$
as desired. 

For more examples, take any $h\in(0,1)$ and let 
$$U_t:=\frac1{\sqrt2}\,(Y_{(1-h)t}+Z_{(1+h)t}),$$
where $(Z_t)$ is an independent copy of the Gaussian process $(Y_t)$. Then 
$$EU_sU_t=1-|t-s|=EY_sY_t \quad\text{if}\quad|t-s|\le1/(1+h)$$
and hence 
$$EU_sU_t=EX_sX_t \quad\text{if}\quad|t-s|\le\min[u,1/(1+h)],$$
as desired.

  [1]: https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)
  [2]: http://www.stat.columbia.edu/~rdavis/papers/VAG002.pdf