Since $y=a^{x}$ and $y=\log _{a}(x+1)$ have the same monotonicity, $f(x)=a^{x}+ \log _{a}(x+1)$ is monotonic on $[0,1]$.

Hence, $f(0)+f(1)=a$, which implies $a^{0}+ \log _{a}1+a^{1}+ \log _{a}2=a$.

Simplifying, we get $1+ \log _{a}2=0$. Solving for $a$, we obtain $a= \frac {1}{2}$.

Therefore, the answer is $\boxed{a= \frac {1}{2}}$.

The monotonicity of the functions $y=a^{x}$ and $y= \log _{a}x$ indicates that $f(x)=a^{x}+ \log _{a}x$ is monotonic on $[0,1]$. Consequently, the maximum and minimum values of the function on $[0,1]$ are $f(0)$ and $f(1)$, respectively. Substituting these values, we can solve for $a$.

This problem primarily assesses the understanding and simple application of the monotonicity of exponential and logarithmic functions. By employing a holistic approach, we can determine the function's extreme values and solve the problem, which is relatively straightforward.