Linear combinations of Poisson random variables
As you've calculated, the moment-generating function of the Poisson distribution with rate $\lambda$ is
$$
m_X(t) = \mathbb E e^{t X} = e^{\lambda (e^t - 1)} \>.
$$
Now, let's focus on a linear combination of independent Poisson random variables $X$ and $Y$. Let $Z = a X + b Y$. Then,
$$
m_Z(t) = \mathbb Ee^{tZ} = \mathbb E e^{t (a X + b Y)} = \mathbb E e^{t(aX)} \mathbb E e^{t (bY)} = m_X(at) m_Y(bt) \>.
$$
So, if $X$ has rate $\lambda_x$ and $Y$ has rate $\lambda_y$, we get
$$
m_Z(t) = \exp({\lambda_x (e^{at} - 1)}) \exp({\lambda_y (e^{bt} - 1)}) = \exp(\lambda_x e^{at} + \lambda_y e^{bt} - (\lambda_x + \lambda_y))\>,
$$
and this cannot, in general, be written in the form $\exp(\lambda(e^t - 1))$ for some $\lambda$ unless $a = b = 1$.
Inversion of moment-generating functions
If the moment generating function exists in a neighborhood of zero, then it also exists as a complex-valued function in an infinite strip around zero. This allows inversion by contour integration to come into play in many cases. Indeed, the Laplace transform $\mathcal L(s) = \mathbb E e^{-s T}$ of a nonnegative random variable $T$ is a common tool in stochastic-process theory, particularly for analyzing stopping times. Note that $\mathcal L(s) = m_T(-s)$ for real valued $s$. You should prove as an exercise that the Laplace transform always exists for $s \geq 0$ for nonnegative random variables.
Inversion can then be accomplished either via the Bromwich integral or the Post inversion formula. A probabilistic interpretation of the latter can be found as an exercise in several classical probability texts.
Though not directly related, you may be interested in the following note as well.
J. H. Curtiss (1942), A note on the theory of moment generating functions, Ann. Math. Stat., vol. 13, no. 4, pp. 430–433.
The associated theory is more commonly developed for characteristic functions since these are fully general: They exist for all distributions without support or moment restrictions.
As mentioned in the comments, characteristic functions always exist, because they require integration of a function of modulus $1$. However, the moment generating function doesn't need to exist because in particular it requires the existence of moments of any order.
When we know that $E[e^{tX}]$ is integrable for all $t$, we can define $g(z):=E[e^{zX}]$ for each complex number $z$. Then we notice that $M_X(t)=g(t)$ and $\varphi_X(t)=g(it)$.
Best Answer
"... why we cannot take $t=−iu$... "
Because, according to the definition of the characteristic function, $t$ must be a real number.