Let $X$ be a nonnegative random variable. We always have $\mathbb E[X] \le \int_0^\infty P(X>x)dx$.
Suppose we also know $P(X>x)\le 100e^{-ax}$. We can integrate naively to see $\mathbb E[X] \le \frac{100}{a}$. We can also be more clever by observing $P(X > x) \le 1$ for all $x$. This gives the bound
$$\mathbb E[X] \le \int_0^\infty P(X>x)dx \le \int_0^\infty \min\{1, 100e^{-ax}\} dx $$ $$ \le \int_0^{\log(100)/a} 1 dx + \int_{\log(100)/a}^\infty 100e^{-ax} dx$$ $$ =\frac{\log 100}{a}+ \frac{100}{a}e^{-ax} \bigg\vert_{x=\log(100)/a}=\frac{1+\log 100}{a}$$
which is much better if $a$ is small.
Now suppose instead we know $P(X > x) \le 100 (e^{-ax}+e^{-bx})$ for some $a,b>0$. We can of course estimate by $P(X > x) \le 200 e^{-\min\{a,b\}x}$ and do the same as above to get $\mathbb E[X] \le \frac{1+\log(100)}{\min\{a,b\}}$.
I wonder is there any way to be more clever than this? There is a temptation to use convexity/concavity to bound $e^{-ax}+e^{-bx}$ by something involving averages. But to my knowledge the convexity/concavity inequalities go the wrong way. Does anyone have an idea?
You can use a very similar approach to what you have already done for $100e^{-ax}$. You have that $$\mathbb{E}[X] = \int_0^{\infty}P(X>x) dx \le \int_0^{\infty} \min\left(1,100\left(e^{-ax}+e^{-bx}\right)\right)$$
Let the solution of $100 \left( e^{-ax}+e^{-bx} \right) = 1$ be denoted as $t$. This means that $$\int_0^{\infty} \min\left(1,100\left(e^{-ax}+e^{-bx}\right)\right) = \int_{0}^{t}1dx+\int_{t}^{\infty}100\left(e^{-ax}+e^{-bx}\right)dx$$
This integral evaluates to $$t + 100\left(\frac{e^{-at}}{a}+\frac{e^{-bt}}{b}\right)$$ Provided you know nothing more about the distribution of $X$, this is the tightest bound you could get.
Because the exponential blows up less than $t$, it is better to have an overestimate of $t$ rather than an underestimate (provided they are off by the same amount). For one approximation, we can use the solution to $2\cdot 100e^{-\min(a, b)x} = 1$. This would be $t\approx t_0=\frac{\ln\left(200\right)}{\min\left(a,b\right)}$. To improve the estimate, you can use the Newton-Raphson method (or any number of root-finding methods).
Specifically, given an estimate $t_n$, a better estimate for $t$ would be $$t_{n+1} = t_{n}+\frac{100\left(e^{-at_{n}}+e^{-bt_{n}}\right)-1}{100\left(ae^{-at_{n}}+be^{-bt_{n}}\right)}$$