My attempt at a solution is as follows:
Let $G$ be a Lie group and $m:G\times G \rightarrow G$ be the multiplication map.
We wish to compute $dm_{(e,e)}: T_{e}G \oplus T_{e}G \rightarrow T_{e}G$. We first compute $dm_{(e,e)}(X,0)$; let $\gamma:(-\epsilon, \epsilon) \rightarrow G \times G$ be a smooth curve such that $\gamma(0)=(e,e)$ and $\gamma'(0)=(X,0)$. If $\gamma(t)=(\gamma_1(t),\gamma_2(t))$ are the component functions, then since $\gamma'(t)=(\gamma'_1(t),\gamma'_2(t))$ it follows that $\gamma_1(0)=\gamma_2(0)=e$ and $\gamma'_1(0)=X$ and $\gamma'_2(0)=0$.
Now, $dm_{(e,e)}(X,0)=(m\circ \gamma)'(0)=(\gamma_1\gamma_2)'(0)=\gamma'_1(0)\gamma_2(0)+\gamma_1(0)\gamma'_2(0)=X$. A similar argument shows $dm_{(e,e)}(0,Y)=Y$. By linearity $dm_{(e,e)}(X,Y)=X+Y$.
I see some people have asked about this problem on SE before, and have come up with slicker solutions that don't rely on the product rule (by taking $\gamma_2(t)=e$). But I am wondering whether the product rule will hold in general in Lie groups as this would be a useful fact.
Yes, it holds on Lie groups. Therefore, you can deduce the formula for $Dm_{(e,e)}$ considering the curve$$\begin{array}{ccc}\mathbb R&\longrightarrow&G\times G\\t&\mapsto&(e^{tX},e^{tY}).\end{array}$$