Every continuous one-parameter subgroup is e^(tA)

Idea. Continuity of $\varphi$ upgrades to smoothness by an averaging trick, after which $\varphi$ solves $\varphi'=A\varphi$.

Step 1: an invertible average. $\varphi$ is continuous and $\varphi(0)=I$. For $c>0$ put $M_c=\int_0^c\varphi(s)\,ds$ (entrywise integral). Then $\frac1cM_c\to\varphi(0)=I$ as $c\to0^+$ (continuity of $\varphi$ at $0$). Since $GL_n$ is open, $\frac1cM_c$ is invertible for $c$ small enough. Fix such a $c$ and write $M=M_c$.

Step 2: differentiability. For every $t$, using the homomorphism property $\varphi(t)\varphi(s)=\varphi(t+s)$,
$$\varphi(t)M=\int_0^c\varphi(t+s)\,ds=\int_t^{t+c}\varphi(u)\,du.$$
The right side is differentiable in $t$ (fundamental theorem of calculus, $\varphi$ continuous), with derivative $\varphi(t+c)-\varphi(t)$. Hence $\varphi(t)=\bigl(\int_t^{t+c}\varphi\bigr)M^{-1}$ is differentiable (even $C^1$).

Step 3: the differential equation. From $\varphi(t+h)=\varphi(t)\varphi(h)$,
$$\frac{\varphi(t+h)-\varphi(t)}h=\varphi(t)\,\frac{\varphi(h)-I}h\ \xrightarrow[h\to0]{}\ \varphi(t)A,\qquad A:=\varphi'(0).$$
So $\varphi'(t)=\varphi(t)A=A\varphi(t)$ (they commute since $\varphi(t)\varphi(h)=\varphi(h)\varphi(t)$).

Step 4: uniqueness for the ODE. Let $\Psi(t)=e^{-tA}\varphi(t)$. Then
$$\Psi'(t)=-Ae^{-tA}\varphi(t)+e^{-tA}A\varphi(t)=0,$$
so $\Psi$ is constant, equal to $\Psi(0)=I$. Therefore $\varphi(t)=e^{tA}$.

Remark. Without continuity there exist wild homomorphisms $\mathbb{R}\to\mathbb{R}^\times$, $t\mapsto e^{f(t)}$, with $f$ an additive non-measurable function (Hamel basis). (It is a classical theorem, proved by a Steinhaus-type argument, that measurable homomorphisms are automatically continuous.)