I know the following is wrong. But I don't see where is wrong. Let's begin by proving $\frac{d}{dt}(A(t)\cdot B(t))=(\frac{d}{dt}A(t))\cdot B(t)+A(t)\cdot\frac{d}{dt}B(t)$, where $A(t)$, $B(t)$ are matrix lie group.
Proof:$$C(t)^i_j=A(t)^i_k\cdot B(t)^k_j,\\ \frac{d}{dt}C(t)^i_j=\frac{d}{dt}(A(t)^i_k\cdot B(t)^k_j=(\frac{d}{dt}A(t)^i_j)\cdot B^k_j+A(t)^i_k\cdot\frac{d}{dt}B(t)^k_j,\\ hence, \ \frac{d}{dt}(A(t)\cdot B(t))=(\frac{d}{dt}A(t))\cdot B(t)+A(t)\cdot\frac{d}{dt}B(t).$$ Then for $e^{tA}$, $e^{tB}$ where $A$, $B$ are two constant matrices, we should have $$\frac{d}{dt}(e^{tA}\cdot e^{tB})=(\frac{d}{dt}e^{tA})\cdot e^{tB}+e^{tA}\cdot\frac{d}{dt}e^{tB}.$$ Letting $t=0$ in this equation, we have $$\frac{d}{dt}|_{t=0}(e^{tA}\cdot e^{tB})=A+B.$$ Then it means that $A+B$ is the generator of $e^{tA}\cdot e^{tB}$. On the other hand, we know $A+B$ generates $e^{t(A+B)}$. So, (by the exponential mapping) we have $e^{tA}\cdot e^{tB}=e^{tA+tB}$ for any matrices $A$ and $B$ which is definitely wrong as known in lie group theory. What's the problem with the analysis? Thank you very much.
If $f(t)=\exp(tC)$ is the $1$-parameter group generated by a matrix $C$, then $f'(t)=Cf(t)$ and $f''(t)=C^2 f(t)$, so that $$f''(0)=f'(0)^2.$$ Let us now consider the function $g(t)=\exp(tA)\exp(tB)$. We have $$g'(t)=A\exp(tA)\exp(tB)+\exp(tA)B\exp(tB)$$ and \begin{align} g''(t)&=A^2\exp(tA)\exp(tB)+A\exp(tA)B\exp(tB)\\ &+A\exp(tA)B\exp(tB)+\exp(tA)B^2\exp(tB). \end{align} It follows that $$g'(0)=A+B$$ as you noticed and that $$g''(0)=A^2+2AB+B^2.$$ If $AB\neq BA$, then $g''(0)\neq g'(0)^2$. This tells us that in that case the function is not a $1$-parameter subgroup.
If instead $AB=BA$ then one can easily check that $g$ is a $1$-parameter subgroup.