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
$$\sigma^2=\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$
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
$${\rm Var}(u_t \mid I_{t-1}) = \omega+\alpha\cdot u_{t-1}^2+\beta\cdot {\rm Var}(u_{t-1} \mid I_{t-2})+\phi\cdot x_{t-1}$$
Taking expectations through out
$$E\big[{\rm Var}(u_t \mid I_{t-1})\big] = \omega+\alpha\cdot E(u_{t-1}^2)+\beta\cdot E\big[{\rm Var}(u_{t-1} \mid I_{t-2})\big]+\phi\cdot E(x_{t-1}) \tag{1}$$
Now, by the Law of Total Variance,
$${\rm Var}(u_t) = E\big[{\rm Var}(u_t \mid I_{t-1})\big] + {\rm{Var}}\big[E(u_t \mid I_{t-1})\big]$$
Under the assumption $E(u_t \mid I_{t-1}) = 0 \implies E(u_t) = 0$, we obtain the relation
$${\rm Var}(u_t) = E\big[{\rm Var}(u_t \mid I_{t-1})\big] \tag{2}$$
and lagging once,
$${\rm Var}(u_{t-1}) = E\big[{\rm Var}(u_{t-1} \mid I_{t-2})\big] \tag{3}$$
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
$${\rm Var}(u) = \omega+\alpha\cdot {\rm Var}(u)+\beta\cdot {\rm Var}(u)+\phi\cdot E(x) $$
$$\implies {\rm Var}(u) =\frac{\omega+\phi\cdot E(x)}{1-\alpha-\beta}$$