For a) to be correct, you'll need your ring to have $1,$ the multiplicative identity. Let me first give you a counter-example to a). Let $R$ be any ring with $1$ and consider $$A:=\left\{\begin{pmatrix}u & v \\ 0 & 0 \end{pmatrix}: \ \ u,v \in R \right\} \subset M_2(R).$$Of course, if you want $A$ to be finite, choose $R$ to be finite, e.g. $R=\mathbb{Z}_2$. Now, let $a=\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \in A.$ Then $s_a$ is injective because $s_a(x)=ax=x$ for all $x \in A.$ But $d_a$ is not injective because, for example, with $x=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix},$ we have $d_a(x)=xa=0.$ But we won't have that problem if we assume that $A$ is finite with $1$. To see that, suppose $s_a$ is injective for some $a \in A$ and $d_a(x)=xa=0$ for some $x \in A.$ Since $A$ is finite, $a^m=a^n$ for some integers $1 \le m < n,$ implying that $a^m(1-a^{n-m})=0$ and hence $1=a^{n-m}$ because $s_a$ is injective. Since $n-m \ge 1,$ we get from $d_a(x)=xa=0$ that $0=xa^{n-m}=x$ and so $d_a$ is injective. Conversely, if $d_a$ is injective, a similar argument works: since $A$ is finite, $a^m=a^n$ for some integers $1 \le m < n$ and so $(1-a^{n-m})a^m=0$ implying that $a^{n-m}=1$ because $d_a$ is injective. So if $s_a(x)=ax=0,$ then $0=a^{n-m}x=x$ proving that $s_a$ is injective.

a) Observe that $s_a$ and $d_a$ are group endomorphisms of $(A,+)$. $s_a$ is injective means that $\ker s_a=\{0\}$ so $a^k$ can never be $0$. Since $A$ is finite, there exists nonnegative integers $i<j$ such that $a^i=a^j$. Hence $aa^{i-1}(1-a^{j-i})=(a^i(1-a^{j-i})=0$ so $a^{i-1}(1-a^{j-i})$ (if $i$ was equal to $0$ we were done before this step). By induction, $a^{j-i}=1$ so $a$ is a unit. Then clearly $\ker d_a=\{0\}$ so $d_a$ is injective. The converse can be proven similarly. $\square$