= Solution
For fixed $u$, write $h(t)=\varphi_{X_t}(u)$. <Independent increments> and <stationary increments> give
$$
h(s+t)=\mathbb E\bigl[e^{iuX_s}e^{iu(X_{s+t}-X_s)}\bigr]
=h(s)h(t),\qquad h(0)=1.
$$
The preceding part gives continuity. Also $h$ never vanishes: if $h(t_0)=0$ with $t_0>0$, then $h(t_0/m)^m=0$ for every positive integer $m$, contradicting $h(t_0/m)\to1$.
Here is a direct proof of the <exponential form of Lévy characteristic functions>. Choose $a>0$ small enough that $|h(t)-1|<1/2$ on $[0,a]$. The principal <complex logarithm> gives a continuous $L(t)=\log h(t)$ there, with $L(0)=0$. For $s,t\geq0$ with $s+t\leq a$, the multiplicative identity implies
$$
L(s+t)-L(s)-L(t)\in2\pi i\mathbb Z.
$$
This difference is continuous on the connected triangle of allowed $(s,t)$ and equals zero at $(0,0)$, so it is identically zero. Thus $L$ satisfies the additive <Cauchy functional equation> locally. Subdivision gives $L(a/m)=L(a)/m$ and $L(ka/m)=kL(a)/m$; continuity then gives $L(t)=tL(a)/a$ for every $t\in[0,a]$.
Define $\eta(u)=L(a)/a$. For any $t\geq0$, choose an integer $m$ with $t/m\leq a$; then
$$
\boxed{\varphi_{X_t}(u)=h(t)=h(t/m)^m=e^{t\eta(u)}.}
$$
The coefficient is unique: if two coefficients give the same exponential for every $t\geq0$, their derivatives at zero agree. In particular $\eta(0)=0$, and $\operatorname{Re}\eta(u)\leq0$ follows from $|h(t)|\leq1$. The <characteristic exponent of a Lévy process> has therefore been obtained from first principles, without invoking the <Lévy–Khintchine formula>.
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