$\mathbf{Note :}$ This is a not so rigorous answer. Here non-rigorous proof of the following identity is given : $$\mathbf{\frac{d}{dt}e_{}^{\hat A(t)}=\int_{0}^{1}dz e_{}^{\hat A(t) z}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]e_{}^{\hat A(t) (1-z)}}.$$
OP's question is about the derivative (with respect to a parameter) of exponential of an operator which depends on a parameter. i.e., $$\frac{d}{dt}e_{}^{\hat A(t)}.$$
This can be deduced using non-commutatitve Liebnitz rule as follows :
(i) Expand the exponential using Taylor-Maclarin formula as : $$\frac{d}{dt}e_{}^{\hat A(t)}= \frac{d}{dt}\sum_{n=0}^{\infty}\frac{1}{\Gamma[n+1]}\hat A(t)_{}^{n}.$$ (ii) Apply non-commutatitve Liebnitz rule to each term of the sum as: $$\frac{d}{dt}e_{}^{\hat A(t)}= \sum_{n=1}^{\infty}\frac{1}{\Gamma[n+1]}\sum_{k=0}^{n-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}.$$ (iii) Use the following identity : $$\frac{1}{\Gamma[k+1]\Gamma[n-k]}\int_{0}^{1}dz z_{}^{k}(1-z_{}^{})_{}^{n-k-1}=\frac{1}{\Gamma[n+1]}$$ to get : $$\frac{d}{dt}e_{}^{\hat A(t)}= \frac{d}{dt}\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{1}{\Gamma[k+1]\Gamma[n-k]}\int_{0}^{1}dz z_{}^{k}(1-z_{}^{})_{}^{n-k-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}$$ which gives : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{1}{\Gamma[k+1]\Gamma[n-k]} z_{}^{k}(1-z_{}^{})_{}^{n-k-1}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k-1}$$ which upon change of summation variable $n \rightarrow n-1$ gives : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{1}{\Gamma[k+1]\Gamma[n-k+1]} z_{}^{k}(1-z_{}^{})_{}^{n-k}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n-k}.$$ (iv) Finally using the change of double summation formula , it follows : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{\Gamma[k+1]\Gamma[n+1]} z_{}^{k}(1-z_{}^{})_{}^{n}\hat A(t)_{}^{k}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]\hat A(t)_{}^{n}.$$ which when rearanged gives : $$\frac{d}{dt}e_{}^{\hat A(t)}= \int_{0}^{1}dz\sum_{k=0}^{\infty}\frac{z_{}^{k}\hat A(t)_{}^{k}}{\Gamma[k+1]} \left[\frac{d}{dt}\hat A(t)_{}^{}\right]\sum_{n=0}^{\infty}\frac{(1-z_{}^{})_{}^{n}\hat A(t)_{}^{k}}{\Gamma[n+1]}.$$ (v) Upon re-exponentiating gives the desired result : $$\mathbf{\frac{d}{dt}e_{}^{\hat A(t)}=\int_{0}^{1}dz e_{}^{\hat A(t) z}\left[\frac{d}{dt}\hat A(t)_{}^{}\right]e_{}^{\hat A(t) (1-z)}}.$$