Andre's solution is great and I do not mean to step on any toes by extending this discussion further however I am a fan of the Poisson distribution and I think the following could be useful if you are going to be solving a lot of Poisson problems.
Another way of solving this problem (and others like it involving the Poisson distribution) is to write $Y = X_1 + X_2$ for the sum of the number of goals in games $1$ and $2$. It turns out that the distribution of $Y$ is also Poisson with parameter $ \lambda_1 + \lambda_2$ where $\lambda_i$ is the parameter of $X_i's$ distribution. In this case we see that the answer is $e^{-8}$ which agrees with Andre's solution. This might not seem like a quicker way to solve the problem, but in general for: $Y_n = \sum_{i=1}^{n}X_i \quad \ \quad X_i \sim Poisson(\lambda_i) \ , X_i \perp X_j \ ;i \neq j$ We get that: $Y_n \sim Poisson(\sum_{i=1}^{n} \lambda_i)$
You may immediately see the usefulness to this if you were instead asked: What is the probability that the same player above scores 15 goals in their first 10 games of the season? Using the above technique you would need to calculate: $\sum_{g_1 +\cdots +g_{10}=15}Pr(X_1=g_1)\cdots Pr(X_{10}=g_{10})$
However with the knowledge of $Y_{10}'s$ distribution we can rattle off the answer quickly as: $\frac{40^{15}e^{-40}}{15!} \approx 3.48\times 10^{-6}$
We can also find all moments, cumulative probabilities, etc. very easily now as well.