1. We are given $f(x+y)=f(x)f'(y)+f'(x)f(y)\,\text{for all }x,y\in\mathbb R,$ with $f(0)=1.$ We need to find $\log_e(f(4))$.

2. Use special values of $x,y$ to derive a differential equation.

Take $x=0$: $f(y)=f(0)f'(y)+f'(0)f(y).$ Since $f(0)=1$, this becomes $f(y)=f'(y)+f'(0)f(y).$ Hence, $f'(y)=(1-f'(0))f(y).$ So $f$ satisfies a first-order linear differential equation of the form $f'(x)=cf(x),$ where $c=1-f'(0).$

3. Now determine $c$ using the given functional equation.

Set $x=0$ and $y=0$ in the original equation: $f(0)=f(0)f'(0)+f'(0)f(0).$ Since $f(0)=1$, $1=1\cdot f'(0)+f'(0)\cdot 1=2f'(0).$ Thus, $f'(0)=\frac12.$ Therefore, $c=1-\frac12=\frac12.$ So $f'(x)=\frac12 f(x).$

4. Solve this differential equation using $f(0)=1$.

The solution of $f'(x)=\frac12 f(x)$ is $f(x)=Ae^{x/2}.$ Using $f(0)=1$ gives $A=1.$ Hence, $f(x)=e^{x/2}.$

5. Verify quickly in the given equation.

We have $f'(x)=\frac12 e^{x/2}.$ Then $f(x)f'(y)+f'(x)f(y)=e^{x/2}\cdot \frac12 e^{y/2}+\frac12 e^{x/2}\cdot e^{y/2}=e^{(x+y)/2}=f(x+y).$ So the function is correct.

6. Compute $f(4)$: $f(4)=e^{4/2}=e^2.$ Therefore, $\log_e(f(4))=\log_e(e^2)=2.$

Thus, the required integer is $\boxed{2}.$