The idea is good. The conditional probabilities are a little bit off: recall that $\Pr(A|B)=\frac{\Pr(A\cap B}{\Pr(B)}$.
So we need to divide each of your terms, and hence your expression, by the probability that $X$ is odd. It is clear that you know how to compute that.
Added: The probability that $X$ is odd is
$$\sum_{i=0}^\infty e^{-\lambda} \frac{\lambda^{2i+1}}{(2i+1)!}.$$
This can be simplified a lot. Write down the power series for $e^\lambda$, also for $e^{-\lambda}$ Subtract. We get twice our sum. So the probability that $X$ is odd is $e^{-\lambda}\left(\frac{e^\lambda-e^{-\lambda}}{2} \right)$. Note this is $e^{-\lambda}\sinh \lambda$.
Now the probability that $X=2k+1$ given that $X$ is odd is, by the usual conditiona probability formula, equal to
$$\frac{e^{-\lambda}\frac{\lambda^{2k+1}}{(2k+1)!}}{e^{-\lambda}\sinh \lambda}.$$
For the conditional expectation, multiply the above expression by $2k+1$, and add up from $k=0$ to $\infty$. Since $\frac{2k+1}{(2k+1)!}=\frac{1}{(2k)!}$, we get, after a little manipulation, that the conditional expectation is
$$\frac{\lambda}{\sinh \lambda}\sum_{k=0}^\infty \frac{\lambda^{2k}}{(2k)!}.$$
By a calculation with the expansions of $e^\lambda$ and $e^{-\lambda}$ of the type we did before, or in another way, we can show that the inner sum is $\cosh \lambda$.
In general, for smooth $g(X)$ you can do a Taylor expansion around the mean $\mu=E(X)$:
$$g(X)=g(\mu) + g'(\mu)(X-\mu)+ \frac{g^{''}(\mu)}{2!}(X-\mu)^2+ \frac{g^{'''}(\mu)}{3!}(X-\mu)^3+\cdots$$
So
$$E[g(X)]=g(\mu) + \frac{g^{''}(\mu)}{2!}m_2+ \frac{g^{'''}(\mu)}{3!} m_3+\cdots $$
where $m_i$ is the $i$-th centered moment. In our case $m_2=m_3 =\lambda$, so:
$$E[g(X)]=\sqrt{\lambda} - \frac{\lambda^{-1/2}}{8} + \frac{ \lambda^{-3/2}}{16} +\cdots $$
This approximation is useful only if $\lambda \gg 1$
Best Answer
EDIT: Sorry about the error, got it corrected now. Thanks, Dilip and Did.
What you need is conditional expectation $\mathbf{E}[X|F]=\frac{E[X \cap F]}{P(F)}$, where $F$ is the set of all odd integers: $$ \mathbf{E}[X \cap F]= \sum_{k=1}^{\infty}\mathbf{E}[X \cap F_k]P[F_k]=1 \cdot e^{-\lambda} \lambda + 3 \cdot \frac{e^{-\lambda}{\lambda^3}}{3!} + \ldots \\ = \sum_{k=0}^{\infty} (2k+1) \frac{e^{-\lambda} \lambda^{2k+1}}{(2k+1)!}=e^{-\lambda} \lambda \sum_{k=1}^{\infty}\frac{\lambda^{2k}}{(2k)!}=e^{-\lambda}\lambda \cosh \lambda $$ At the same time. $P(F)=e^{-\lambda} \sinh \lambda$. Hence, $\mathbf{E}[X|F]=\frac{\mathbf{E}[X \cap F]}{P(F)} = \lambda \coth \lambda$