You don't need to worry about expectations and covariances, since the difference operator kills the polynomial term. In other words, the result turns on what the difference operator does to polynomials. Once you have taken $d$ differences, you are only left with the stochastic part of your model.
So let's prove this. Since the difference operator is linear, you only have to prove the result for $t^d$.
I would go for a proof by induction. The first difference kills the linear term, since you get $at-b - a(t-1)-b=2b$ A first difference gives you a stationary time series with non-zero mean.
For higher polynomials, say $d$, the first difference gives $t^d-(t-1)^d=t^d-t^d -nt^{n-1}+\text{a polynomial of degree}\ d-1$.
See what happens when we take the second difference. We have $nt^{n-1}$ (plus lower order junk) against $n(t-1)^{n-1}$ (plus lower order junk) from the difference of the next two terms. The induction assumption works here, because we have the same coefficient for the leading term. Note that we are making use of the assumption that the time series is sampled at equally spaced moments, or we would have got a different leading term in the second "first difference". We usually make this assumption for time series, and you can see that it matters in this situation.
You can argue inductively for both the necessary and the sufficient condition.
You still have to deal with $X_t$, the stationary part of the expression. You need to show that the first difference of a stationary time series is also stationary - but that's obvious.You only need to show this for the first difference.
There are several different flavours of stationarity. The type described in your definition is weak-sense stationarity, also known as wide-sense stationarity, covariance stationarity, or second-order stationarity.
Your definition is not quite complete: a preliminary condition for weak stationarity is that the mean and covariance must exist and be finite, but this is satisfied here. As you noted, weak stationarity further requires the mean to be constant over time and $\operatorname{Cov}(X_{t+h},X_t)$ to be independent of $t$ for each $h$ i.e. the autocovariance at each lag $h$ is constant over time. The fact that $\mathbb{E}(t)=a+bt$ shows the first of these conditions is not met. Even the mean is not stationary.
Another form of stationarity is strong stationarity, also called strict stationarity or just stationarity. This requires the joint distribution function of the joint distribution of $X_t$ taken at any $k$ times $t_1, t_2, \dots , t_k$ is the same when lagged by any $\tau$. Technically, for any $k$ and any $\tau$, and for any $t_1, t_2, \dots, t_k$ we require
$$F_X(x_{t_1+\tau}, \dots, x_{t_k+\tau}) = F_X(x_{t_1}, \dots, x_{t_k})$$
This does not just imply that the mean and covariance at any given lag (if either exists) must stay constant over time, but that every conceivable property one can derive from the distribution is invariant under a time shift, in which sense its conditions are "stricter" (though note that if mean or covariance are not finite, we can have a strongly stationary series which does not fulfil the preliminary condition for weak stationarity). Since your mean is not constant over time, then your process can't be strongly stationary either. In general, so long as the mean and covariance exist, we can conclude that a process that is not weakly stationary will not be strongly stationary either. The converse is not true: just because a process is not strongly stationary, doesn't mean it can't be weakly stationary, since it is possible to be weakly stationary yet not strongly stationary.
There is a sense in which your $X_t$ is "stationary": it is trend stationary.
This means that the trend in your time series can be expressed as a function of $t$; if we strip this trend out then what we are left with is a stationary process. In particular, $X_t$ is trend stationary if we can express it as
$$X_t = f(t) + Y_t$$
where $f$ is a deterministic function of time $t$ and $\{Y_t\}$ is a stationary process. In your case we can take $f(t)=a + bt$ and $Y_t = Z_t$; the idea is that by "stripping out" the trend $f(t)$ we would have $X_t - (a + bt) = Z_t$ which is stationary (because it's white noise).
Alternatively we could have taken $f(t)=bt$ and $Y_t= a + Z_t$.
Note that although in this case we could take a process that didn't have a constant mean, then can subtract a deterministic trend to obtain a stationary process, doesn't mean we can do this on any such process. Consider a random walk with drift, for instance. Here the mean is not constant over time, but we can't simply subtract a deterministic trend to obtain a stationary process. So a random walk with drift would not be a trend stationary process. See the question "Difference between series with drift and series with trend" for an illustration.
Best Answer
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.