Let $E$ be a $\mathbb R$-Banach space and $\varphi\in C^0(E,\mathbb C\setminus\{0\})$.
Let $x\in E$. How can we show that "there is an unique branch $h(t)$, $0\le t\le 1$, of $\ln\varphi(tx)$ such that $h(0)=0$ and $h$ is continuous" and what's the precise meaning of this claim?
It' clear to me that if $z\in\mathbb C\setminus\{0\}$, then $\phi\in\mathbb R$ is called argument of $z$ if $$z=|z|(\cos\phi+{\rm i}\sin\phi)\tag1.$$ If $\operatorname{arg}z$ denotes the set of all arguments of $z$, then the unique element $\operatorname{Arg}z\in\operatorname{arg}z\cap(-\pi,\pi]$ is called the *principal value of the argument of $z$ and we define $$\ln z:=\ln|z|+{\rm i}\operatorname{Arg}z\tag2.$$
I guess the main issue we deal with in the question is the lack of continuity of $[0,1]\ni t\mapsto\operatorname{Arg}\varphi(tx)$.
This is a classic result and one proof could go like this: first note that $t \to \phi(tx)$ is a continuous function from $[0,1] \to \mathbb C^*$ so the space $E$ is irrelevant for our purposes and we can just consider the case $t \to z(t)$ a continuous function from $[0,1] \to \mathbb C^*$.
Since the image is compact and avoids zero, there is $\delta >0, |z(t)| \ge \delta, t \in [0,1]$ and since the function is uniformly continuous, there is $\lambda>0, |z(t)-z(s)| < \delta, |t-s| < \lambda$ and we can choose a partition $0=t_0<t_1<...t_m=1, t_{k+1}-t_{k} < \lambda$
Consider $z(t)=|z(t)|e^{\psi(t)}$ where $\psi(t)$ is the principal argument in $-\pi <\psi(t) \le \pi$ and we construct by induction $\phi(t)$ continuous argument.
We define $\phi(a)=\psi(a)$ and if $\pi \notin \psi [a,t_1]$ (equivalently $z(t) \notin (-\infty, 0)$ there) then $\psi$ is continuous there (the only discontinuity is at $\pi$ by our choice) so we don't change anything and let $\phi=\psi$ there.
If $\pi \in \psi [a,t_1], \psi(s)=\pi$ we claim that there is no $s_1 \in [a,t_1], \psi(s_1)=0$ as otherwise $\delta > |z(s_1)-z(s)|=|z(s_1)|+|z(s)| > |z(s_1)| > \delta$ by our choices which is impossible. This means that we can take a continuous argument $\psi_1(t) \in (0, 2\pi)$ now there and if $\psi(a)-\psi_1(a)=2m_0\pi$ we define $\phi(t)=\psi_1(t)+2m_0\pi$, $\phi(a)=\psi(a)$ as required.
But now it is clear that we can repeat this on $[t_1,t_2]$ with $t_1=a$ now and find $\phi$ continuous argument, where the only change now is that in either case ($\psi$ continuous on $[t_1,t_2]$ or $\psi_1$ with values in $(0, 2\pi)$ continuous argument there), we shift by the appropriate integer multiple of $2\pi$ to preserve the value $\phi(t_1)$ and keep it continuous from the left there too, so by induction we can continue until we reach $t_k=1$ and we are done and can find $\phi$ continuous argument!