$\mathbf{Note :}$ This is a not so rigorous answer.

OP's question is about the derivative (with respect to a parameter) of exponential of an operator which depends on a parameter. 
i.e., $$\hat O(t)= \frac{d}{dt}e_{}^{\hat A(t)}.$$

This can be deduced using [non-commutatitve Liebnitz rule](https://physics.stackexchange.com/questions/359163/time-dependence-of-a-function-of-an-operator/359236#359236) as follows :

(i) Expand the exponential using Taylor-Maclarin formula as :
$$\hat O(t)= \frac{d}{dt}\sum_{n=0}^{\infty}\frac{1}{\Gamma[n+1]}\hat A(t)_{}^{n}.$$
(ii) Apply [non-commutatitve Liebnitz rule](https://physics.stackexchange.com/questions/359163/time-dependence-of-a-function-of-an-operator/359236#359236) to each term of the sum as:
$$\hat O(t)= \sum_{n=1}^{\infty}\frac{1}{\Gamma[n+1]}\sum_{k=0}^{n-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}.$$
(iii) Use the following [identity](https://dlmf.nist.gov/5.12) :
$$\frac{1}{\Gamma[k+1]\Gamma[n-k]}\int_{0}^{1}dz z_{}^{k}(1-z_{}^{})_{}^{n-k-1}=\frac{1}{\Gamma[n+1]}$$ 
to get :
$$\hat O(t)= \frac{d}{dt}\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{1}{\Gamma[k+1]\Gamma[n-k]}\int_{0}^{1}dz z_{}^{k}(1-z_{}^{})_{}^{n-k-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}$$
which gives :
$$\hat O(t)= \int_{0}^{1}dz\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{1}{\Gamma[k+1]\Gamma[n-k]} z_{}^{k}(1-z_{}^{})_{}^{n-k-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}$$ which upon change of summation variable $n \rightarrow n-1$ gives :
$$\hat O(t)= \int_{0}^{1}dz\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{1}{\Gamma[k+1]\Gamma[n-k+1]} z_{}^{k}(1-z_{}^{})_{}^{n-k}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k}.$$
(iv) Finally using the [change of double summation formula](http://functions.wolfram.com/GeneralIdentities/12/) , it follows :
$$\hat O(t)= \int_{0}^{1}dz\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{\Gamma[k+1]\Gamma[n+1]} z_{}^{k}(1-z_{}^{})_{}^{n}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n}.$$
which when rearanged gives :
$$\hat O(t)= \int_{0}^{1}dz\sum_{k=0}^{\infty}\frac{z_{}^{k}\hat A(t)_{}^{k}}{\Gamma[k+1]} \left[\frac{d}{dt}\hat A(t)_{}^{}\right]\sum_{n=0}^{\infty}\frac{(1-z_{}^{})_{}^{n}\hat A(t)_{}^{k}}{\Gamma[n+1]}.$$
(v) Upon re-exponentiating gives the desired result :
$$\mathbf{\frac{d}{dt}e_{}^{\hat A(t)}=\int_{0}^{1}dz e_{}^{\hat A(t) z}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]e_{}^{\hat A(t) (1-z)}}.$$