Perhaps surprisingly, this is not true. (Independence of the two time series will make it true, however.)

I understand "stable" to mean stationary, because those words appear to be used interchangeably in millions of search hits, including at least one on our site.

For a counterexample, let $X$ be a non-constant stationary time series for which every $X_t$ is independent of $X_s$, $s\ne t,$ and whose marginal distributions are symmetric around $0$. Define

$$Y_t = (-1)^t X_t.$$

These plots show portions of the three time series discussed in this post. $X$ was simulated as a series of independent draws from a standard Normal distribution.

To show that $Y$ is stationary, we need to demonstrate that the joint distribution of $(Y_{s+t_1}, Y_{s+t_2}, \ldots, Y_{s+t_n})$ for any $t_1\lt t_2 \lt \cdots \lt t_n$ does not depend on $s$. But this follows directly from the symmetry and independence of the $X_t$.

These lagged scatterplots (for a sequence of 512 values of $Y$) illustrate the assertion that the joint bivariate distributions of $Y$ are as expected: independent and symmetric. (A "lagged scatterplot" displays the values of $Y_{t+s}$ against $Y_{t}$; values of $s=0,1,2$ are shown.)

Nevertheless, choosing $\alpha=\beta=1/2$, we have

$$\alpha X_t + \beta Y_t = X_t$$

for even $t$ and otherwise

$$\alpha X_t + \beta Y_t = 0.$$

Since $X$ is non-constant, obviously these two expressions have different distributions for any $t$ and $t+1$, whence the series $(X+Y)/2$ is not stationary. (It's not even first-order stationary.) The colors in the first figure highlight this non-stationarity in $(X+Y)/2$ by distinguishing the zero values from the rest.

Perhaps surprisingly, this is not true. (Independence of the two time series will make it true, however.)

I understand "stable" to mean stationary, because those words appear to be used interchangeably in millions of search hits, including at least one on our site.

For a counterexample, let $X$ be a non-constant stationary time series for which every $X_t$ is independent of $X_s$, $s\ne t,$ and whose marginal distributions are symmetric around $0$. Define

$$Y_t = (-1)^t X_t.$$

These plots show portions of the three time series discussed in this post. $X$ was simulated as a series of independent draws from a standard Normal distribution.

To show that $Y$ is stationary, we need to demonstrate that the joint distribution of $(Y_{s+t_1}, Y_{s+t_2}, \ldots, Y_{s+t_n})$ for any $t_1\lt t_2 \lt \cdots \lt t_n$ does not depend on $s$. But this follows directly from the symmetry and independence of the $X_t$.

These lagged scatterplots (for a sequence of 512 values of $Y$) illustrate the assertion that the joint bivariate distributions of $Y$ are as expected: independent and symmetric. (A "lagged scatterplot" displays the values of $Y_{t+s}$ against $Y_{t}$; values of $s=0,1,2$ are shown.)

Nevertheless, choosing $\alpha=\beta=1/2$, we have

$$\alpha X_t + \beta Y_t = X_t$$

for even $t$ and otherwise

$$\alpha X_t + \beta Y_t = 0.$$

Since $X$ is non-constant, obviously these two expressions have different distributions for any $t$ and $t+1$, whence the series $(X+Y)/2$ is not stationary. The colors in the first figure highlight this non-stationarity in $(X+Y)/2$ by distinguishing the zero values from the rest.