How does explanatory variables in variance affect unconditional variance of GARCH(1,1)

Your last expression is not correct because, as noted in the comments, you are after the unconditional variance, which is constant in these models. it should be

$$sigma^2=frac{omega+phicdot E(x)}{1-alpha-beta}$$

PS: Also, you are missing a $t$-subscript in your penultimate expression.

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RESPONSE to comment
Using more explicit notation, we assume that the conditional variance is

$${rm Var}(u_t mid I_{t-1}) = omega+alphacdot u_{t-1}^2+betacdot {rm Var}(u_{t-1} mid I_{t-2})+phicdot x_{t-1}$$

Taking expectations through out

$$Ebig[{rm Var}(u_t mid I_{t-1})big] = omega+alphacdot E(u_{t-1}^2)+betacdot Ebig[{rm Var}(u_{t-1} mid I_{t-2})big]+phicdot E(x_{t-1}) tag{1}$$

Now, by the Law of Total Variance,

$${rm Var}(u_t) = Ebig[{rm Var}(u_t mid I_{t-1})big] + {rm{Var}}big[E(u_t mid I_{t-1})big]$$

Under the assumption $E(u_t mid I_{t-1}) = 0 implies E(u_t) = 0$, we obtain the relation

$${rm Var}(u_t) = Ebig[{rm Var}(u_t mid I_{t-1})big] tag{2}$$

and lagging once,

$${rm Var}(u_{t-1}) = Ebig[{rm Var}(u_{t-1} mid I_{t-2})big] tag{3}$$

Another assumption is that $u$ is homoskedastic unconditionally. This together with $E(u_t)=0$ means

$${rm Var}(u_t) = {rm Var}(u_{t-1}) = E(u^2) tag{4}$$

Finally, another assumption is that the $x$'s are identically distributed over time, so $E(x_{t-1}) = E(x)$.

Using all these into $(1)$ we get

$${rm Var}(u) = omega+alphacdot {rm Var}(u)+betacdot {rm Var}(u)+phicdot E(x) $$

$$implies {rm Var}(u) =frac{omega+phicdot E(x)}{1-alpha-beta}$$