Let $B$ be the indicator of what type of jump occurs, if any; $B=-1$ when there's a negative jump, $B=1$ when there's a positive jump and $B=0$ if there is no jump. Then to determine the jump, you just need to generate $B$. According to your formulation,

$$ P(B=-1) = p_{down} $$

$$P(B=1) = p_{up} $$

$$P(B=0) = 1-p_{up}-p_{down} $$

You can generate $B$ by generating an Uniform(0,1) variable $U$ and letting

$$B=-1 \ \ \ {\rm if} \ \ \ U \leq p_{down}, $$

$$B=1 \ \ \ {\rm if} \ \ \ p_{down} \leq U \leq p_{down}+p_{up} $$

and

$$B=0 \ \ \ {\rm if} \ \ \ U \geq p_{down}+p_{up}$$

Examining the Uniform(0,1) CDF will make it clear why the probabilities work out correctly.

You could integrate out $Y$ if the question were to determine the distribution of $S(T)$ analytically but I think the point is to determine its distribution by simulation.

Note: Assuming you don't have a function to generate normally distributed variables, you will also have to generate more uniforms to simulate $X$. If you use the Box-Muller transform method of generating normals, you can simulate two normals for every two uniformly distributed variables.

Let $B$ be the indicator of what type of jump occurs, if any; $B=-1$ when there's a negative jump, $B=1$ when there's a positive jump and $B=0$ if there is no jump. Then to determine the jump, you just need to generate $B$. According to your formulation,

$$ P(B=-1) = p_{down} $$

$$P(B=1) = p_{up} $$

$$P(B=0) = 1-p_{up}-p_{down} $$

You can generate $B$ by generating a Uniform(0,1) variable $U$ and letting

$$B=-1 \ \ \ {\rm if} \ \ \ U \leq p_{down}, $$

$$B=1 \ \ \ {\rm if} \ \ \ p_{down} \leq U \leq p_{down}+p_{up} $$

and

$$B=0 \ \ \ {\rm if} \ \ \ U \geq p_{down}+p_{up}$$

Examining the Uniform(0,1) CDF will make it clear why the probabilities work out correctly.

You could integrate out $Y$ if the question were to determine the distribution of $S(T)$ analytically but I think the point is to determine its distribution by simulation.

Note: Assuming you don't have a function to generate normally distributed variables, you will also have to generate more uniforms to simulate $X$. If you use the Box-Muller transform method of generating normals, you can simulate two normals for every two uniformly distributed variables.