Polynomial-Time reduction from Partition to MakeSpan
Given an instance of partition (i.e., a set of numbers) $\{a_1, \dots, a_n\}$ create an instance of Job Scheduling (what you call Makespan) with $2$ machines and $n$ jobs $j_1, \dots, j_n$, where the execution time of the $i$-th job is $a_i$.
Pick $b = \frac{1}{2} \sum_{i=1}^n a_i$.

If there is a solution to the partition problem, i.e., a set $A \subseteq \{a_1, \dots, a_n\}$ such that $\sum_{a \in A} a = \frac{1}{2} \sum_{i=1}^n a_i$, then there is a solution for the job scheduling problem: simply assign $j_i$ to machine $1$ if $a_i \in A$, and to machine 2 otherwise.

If there is a solution to the job scheduling problem, i.e., a partition of jobs into $M_1$ and $M_2$ such that $\max\{ \sum_{j_i \in M_1} a_i, \sum_{j_i \in M_2} a_i \} \le b$ then there is a solution $A$ to the partition problem.
The choice of $b$ implies $\sum_{j_i \in M_1} a_i = b = \frac{1}{2} \sum_{i=1}^n a_i$, therefore it suffices to select $A = \{a_i \, : \; j_i \in M_1 \}$.