[proofplan] We prove the two implications directly from the definition of an $R$-submodule. If $N$ is a submodule, then scalar multiples of elements of $N$ remain in $N$, and the sum of two such scalar multiples remains in $N$. Conversely, the assumed closure under all two-term $R$-linear combinations recovers closure under addition, additive inverses, and scalar multiplication by choosing the coefficients $1_R$, $-1_R$, and $0_R$ appropriately. [/proofplan]

[step:Use submodule closure to obtain every two-term linear combination] Assume that $N$ is an $R$-submodule of $M$. Let $a,b \in R$ and let $x,y \in N$. Since $N$ is closed under the scalar action of $R$, both $ax$ and $by$ belong to $N$. Since $N$ is closed under the addition inherited from $M$, their sum also belongs to $N$: \begin{align*} ax + by \in N. \end{align*} Thus every two-term $R$-linear combination of elements of $N$ lies in $N$. [/step]

[step:Recover the submodule operations from the linear combination hypothesis]Assume conversely that for every $a,b \in R$ and every $x,y \in N$, \begin{align*} ax + by \in N. \end{align*} Because $N$ is nonempty, choose an element $y_0 \in N$. First, let $x,y \in N$. Taking $a = 1_R$ and $b = 1_R$ in the hypothesis gives \begin{align*} 1_R x + 1_R y \in N. \end{align*} Since $M$ is a unital left $R$-module, $1_R x = x$ and $1_R y = y$, hence \begin{align*} x+y \in N. \end{align*} Therefore $N$ is closed under addition. Next, let $x \in N$. Taking $a = -1_R$, $b = 0_R$, and $y = y_0$ in the hypothesis gives \begin{align*} (-1_R)x + 0_R y_0 \in N. \end{align*} The module axioms give $0_R y_0 = 0_M$ and $(-1_R)x = -x$, so \begin{align*} -x \in N. \end{align*} Therefore $N$ is closed under additive inverses. Finally, let $r \in R$ and $x \in N$. Taking $a = r$, $b = 0_R$, and $y = y_0$ in the hypothesis gives \begin{align*} rx + 0_R y_0 \in N. \end{align*} Since $0_R y_0 = 0_M$, this says \begin{align*} rx \in N. \end{align*} Thus $N$ is closed under scalar multiplication by elements of $R$.[/step]

[guided]The converse is the part where the two-term linear combination condition must be unpacked carefully. We assume that for every $a,b \in R$ and every $x,y \in N$, \begin{align*} ax + by \in N. \end{align*} The subset $N$ is nonempty, so we may fix one element $y_0 \in N$. This fixed element is used whenever we want to turn the two-variable condition into a one-variable closure statement. To prove closure under addition, take arbitrary $x,y \in N$. We choose the coefficients $a = 1_R$ and $b = 1_R$. The hypothesis gives \begin{align*} 1_R x + 1_R y \in N. \end{align*} Because $M$ is a unital left $R$-module, the identity element $1_R$ acts as the identity on $M$. Hence $1_R x = x$ and $1_R y = y$, so the displayed element is exactly $x+y$. Therefore \begin{align*} x+y \in N. \end{align*} This proves that $N$ is closed under addition. To prove closure under additive inverses, take arbitrary $x \in N$. We must produce $-x$ as a permitted linear combination of elements of $N$. Since the condition expects two elements of $N$, we use the fixed element $y_0 \in N$ for the second input and choose coefficients $a = -1_R$ and $b = 0_R$. The hypothesis gives \begin{align*} (-1_R)x + 0_R y_0 \in N. \end{align*} The module axioms give $0_R y_0 = 0_M$, and the scalar action of the additive inverse $-1_R$ gives $(-1_R)x = -x$. Thus the displayed element is $-x$, and therefore \begin{align*} -x \in N. \end{align*} This proves that $N$ is closed under additive inverses. To prove closure under scalar multiplication, take arbitrary $r \in R$ and $x \in N$. Again we use the fixed element $y_0 \in N$ for the second input, this time with coefficients $a = r$ and $b = 0_R$. The hypothesis gives \begin{align*} rx + 0_R y_0 \in N. \end{align*} Since $0_R y_0 = 0_M$, this reduces to \begin{align*} rx \in N. \end{align*} Thus $N$ is closed under scalar multiplication by every element of $R$.[/guided]

[step:Conclude that the subset is exactly a submodule] From the previous step, $N$ is nonempty, closed under addition, closed under additive inverses, and closed under scalar multiplication by elements of $R$. Therefore $N$ is an additive subgroup of the underlying abelian group of $M$ and is stable under the $R$-action. By the definition of an $R$-submodule, $N \le M$. This proves the converse implication and completes the proof. [/step]