Consider a distribution with
$$P(X=m\delta)=1-\delta$$
$$P\left(X=m\tfrac{1-\delta+\delta^2}{\delta}\right)=\delta$$
then $E[X]=m$ but $P(X\gt a )$ can be made arbitrarily small by choosing $\delta \lt \frac{a}{m}$ and if necessarily smaller.
So to get something useful, you would need some constraint on $X$ such as a given maximum or a given variance.
Lower bound
Since the event in Chebyshev's inequality
$$P(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2}\tag{*}\label{*}$$
is symmetric about $\mu = E(X) = 10$, so we consider $P(0 < X < 20) \le P(0 < X < 40)$ and try to find a lower bound for this question. Now we take $\epsilon = 10, \mu = 10, \sigma = 4$ in \eqref{*}.
\begin{aligned}
P(|X - 10| \ge 10) &\le \left(\frac{4}{10}\right)^2 \\
P(X \le 0 \text{ or } X \ge 20) &\le \frac{4}{25} \\
P(0 < X < 20) &= P(|X-10|<10) \\
&= 1-P(|X-10|\ge10) \\
&\ge 1 - \frac{4}{25} = \frac{21}{25}
\end{aligned}
Hence a lower bound would be
$$\frac{21}{25} \le P(0 < X < 20) \le P(0 < X < 40).$$
Upper bound
Without further information about random variable $X$, we can only conclude that $P(0 < X < 40)$ is bounded above by one by definition of probability due to the existence of a random variable which is distributed near enough to its mean.
Consider a random variable
$$X =
\begin{cases}
14 & \text{ with probability } \frac12 \\
6 & \text{ with probability } \frac12
\end{cases}
$$
It's clear that $\mu = E(X) = 10$ and
\begin{align}
\sigma =& \sqrt\frac{(14-10)^2+(6-10)^2}{2} \\
=& \sqrt\frac{4^2 + 4^2}{2} = 4.
\end{align}
So $X$ satisfies the hypothesis of the question.
Therefore, in this particular case, $P(0 < X < 40) = 1$. In other words, any number less than one won't be an upper bound to the required probability.
Conclusion
In general, we have $$\frac{21}{25} \le P(0 < X < 40) \le 1.$$
Best Answer
Your bound is wrong since Markov's inequality assumes that $\mathbb{P}(X<0)=0.$ See the proof in Wikipedia for example. This is also why the $\mathbb{P}(X<a)=0$ never entered the picture.
Apply Markov to $Y=X+a,$ which obeys the assumption $\mathbb{P}(Y<0)=0$. This will give $$ 1-\mathbb{P}(Y<b) \leq \frac{\mathbb{E}(Y)}{b}=\frac{a}{b}, $$ leading to the lower bound $$ 1+ \frac{a}{b} \leq \mathbb{P}(Y<b) , $$ which is in $(0,1)$ since $b<0.$ Rewritten this gives $$ 1+ \frac{a}{b} \leq \mathbb{P}(X<b-a). $$