This conditional expectation can be directly evaluated by the formula $E[\xi|A] = \frac{E[\xi I_A]}{P(A)}$ with $P(A) > 0$.

Since $X \sim \text{Poisson}(\lambda c)$ and $Y \sim \text{Poission}(\lambda(1 - c))$ are independent, it follows that $X + Y \sim \text{Poission}(\lambda)$, whence \begin{align*} P(X + Y > 0) = 1 - P(X + Y = 0) = 1 - e^{-\lambda}. \end{align*}

On the other hand, \begin{align*} & E\left[\frac{X}{X + Y}I_{\{1, 2, 3, \ldots\}}(X + Y)\right] \\ =& \sum_{k = 1}^\infty E\left[\frac{X}{X + Y}I_{\{k\}}(X + Y)\right] \\ =& \sum_{k = 1}^\infty k^{-1}E\left[XI_{\{k\}}(X + Y)\right] \\ =& \sum_{k = 1}^\infty k^{-1}\sum_{\substack{m \geq 0, n \geq 0 \\ m + n = k}}mP(X = m, Y = n) \\ =& \sum_{k = 1}^\infty k^{-1}\sum_{\substack{m \geq 0, n \geq 0 \\ m + n = k}}mP(X = m)P(Y = n) \\ =& \sum_{k = 1}^\infty k^{-1}\sum_{\substack{m \geq 0, n \geq 0 \\ m + n = k}}me^{-\lambda c}\frac{(\lambda c)^m}{m!}e^{-\lambda(1 - c)}\frac{(\lambda(1 - c))^n}{n!} \\ =& \sum_{k = 1}^\infty k^{-1}\sum_{m = 0}^kme^{-\lambda c}\frac{(\lambda c)^m}{m!}e^{-\lambda(1 - c)}\frac{(\lambda(1 - c))^{k - m}}{(k - m)!} \\ =& \sum_{k = 1}^\infty k^{-1}e^{-\lambda}\lambda^k\sum_{m = 1}^k\frac{1}{(m - 1)!(k - m)!}c^m(1 - c)^{k - m} \\ =& \sum_{k = 1}^\infty k^{-1}e^{-\lambda}\lambda^k\frac{1}{(k - 1)!}c\sum_{l = 0}^{k - 1}\binom{k - 1}{l}c^l(1 - c)^{k - 1 - l} \\ =& c\sum_{k = 1}^\infty \frac{1}{k!}e^{-\lambda}\lambda^k \\ =& c(1 - e^{-\lambda}). \end{align*} Therefore, \begin{align*} E\left[\left.\frac{X}{X + Y} \right| X + Y > 0\right] = \frac{E\left[\frac{X}{X + Y}I_{\{1, 2, 3, \ldots\}}(X + Y)\right]}{P(X + Y > 0)} = \frac{c(1 - e^{-\lambda})}{1 - e^{-\lambda}} = c. \end{align*}