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The printed pathwise conclusion is false without specifying the initial value. We first prove exactly what the hypotheses do imply, by stationary drift saturation.

Stationarity gives $\mathbb{E}{Z}_s^2=1$. The Itô formula yieldsThe stochastic integral is a true square-integrable martingale on every finite horizon, since $\mathbb{E}{\int }_0^t{Z}_s^{2}ds=t$. Also $\mathbb{E}|Z_s{g}_s|\le \lambda$ by the Cauchy-Schwarz inequality. Taking expectations therefore givesEvery integrand is nonnegative, so $\mathbb{E}[Z_s{g}_s]=-\lambda$ for almost every $s$. For those times,It follows that $g_s=-\lambda Z_s$ for almost every time and outcome. Equality of these integrable drift terms gives the exact solutionThus $Z$ solves the same equation, but stationarity has not identified its initial random variable with $B_1$.

For an explicit counterexample, takeThis is adapted to ${\mathcal{G}}_t$. Its independent normal initial value and Brownian noise make it a centered stationary Gaussian process with covariance $e^{-\lambda |t-s|}$, exactly as for $Y$. Hence $\mathbb{E}{Z}_0^2=1$ and $\mathbb{E}{g}_t^2={\lambda }^2$, satisfying every printed hypothesis. Nevertheless $Z_t-Y_t=-2e^{-\lambda t}{B}_1$, which is nonzero almost surely for every finite $t$.

The corrected result is indistinguishability if one adds $Z_0=Y_0$ almost surely. Without that addition, there is equality in law of the stationary Ornstein-Uhlenbeck processes, not necessarily equality of their coupled paths. To see that the initial law is forced, independence of Brownian increments from ${\mathcal{G}}_0$ and stationarity give, for its characteristic function $\phi$,Letting $t\to \infty$ proves $\phi (u)=e^{-u^2/2}$. This is the distinction captured by same-noise stationary Ornstein-Uhlenbeck processes.