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Here is how I would start.

Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that $M \otimes N^* \to \hom(N,M)$, $m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$ is a monomorphism. Both sides are additive functors in both variables. It follows formally:

1. The class $S$ is closed under finite direct sums (in both variables).
2. The class $S$ is closed under direct summands (in both variables).

Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).

We can also prove:

3. If $I$ is any index set and $(M_i,N) \in S$ for $i \in I$, then $(\bigoplus_{i \in I} M_i,N) \in S$:

This is because $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i)$ is mono iff the composition $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i) \hookrightarrow \hom(N,\prod_{i \in I} M_i)$ is mono, which identifies with $\bigoplus_i (M_i \otimes N) \hookrightarrow \bigoplus_{i \in I} \hom(N,M_i) \hookrightarrow \prod_{i \in I} \hom(N,M_i)$.

In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)

The next case would be to look at two cyclic modules, say $M = A/I$ and $N = A/J$ for two ideals $I,J$. Then $M \otimes N^* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$ and $\hom(N,M) \cong (I:J)/I$, and the canonical map is given by $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $\mathrm{Ann}(I) \to A/I$, $a \mapsto [a]$ with kernel $I \cap \mathrm{Ann}(I)$. This kernel can be non-trivial: take $A = k[X]/\langle X^2 \rangle$ and $I = \langle X \rangle$.

Here is how I would start.

Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that the canonical map $$M \otimes N^* \to \hom(N,M), ~ m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$$ is a monomorphism. Both sides are additive functors in both variables. It follows formally that $S$ is closed under finite direct sums (in both variables) as well as under direct summands (in both variables).

Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).

We can also prove that $S$ is closed under infinite direct sums in the first variable: Let $I$ be any index set and $(M_i,N) \in S$ for $i \in I$, then $(\bigoplus_{i \in I} M_i,N) \in S$: This is because the map $$\textstyle (\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i)$$ is mono iff the composition $$\textstyle (\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i) \hookrightarrow \hom(N,\prod_{i \in I} M_i)$$ is mono, which identifies with the mono $$\textstyle \bigoplus_i (M_i \otimes N) \hookrightarrow \bigoplus_{i \in I} \hom(N,M_i) \hookrightarrow \prod_{i \in I} \hom(N,M_i).$$

In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)

Remark: $S$ is not closed under infinite direct sums in the second variable. In fact, it can happen that $M \otimes A^I \to M^I$ is not a mono, so that $(M,\bigoplus_{i \in I} A) \notin S$.

The next case would be to look at two cyclic modules, say $$M = A/I,~N = A/J$$ for two ideals $I,J$. Then $$M \otimes N^* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$$ and $$\hom(N,M) \cong \{[a] \in A/I : aJ \subseteq I\} = (I:J)/I.$$ The canonical map then identifies with $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $$\mathrm{Ann}(I) \to A/I,~a \mapsto [a]$$ with kernel $I \cap \mathrm{Ann}(I)$. This kernel can be non-trivial: take $A = k[X]/\langle X^2 \rangle$ and $I = \langle X \rangle$.

Here is how I would start.

Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that $M \otimes N^* \to \hom(N,M)$, $m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$ is a monomorphism. Both sides are additive functors in both variables. It follows formally:

1. The class $S$ is closed under finite direct sums (in both variables).
2. The class $S$ is closed under direct summands (in both variables).

Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).

We can also prove:

3. If $I$ is any index set and $(M_i,N) \in S$ for $i \in I$, then $(\bigoplus_{i \in I} M_i,N) \in S$:

This is because $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i)$ is mono iff the composition $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i) \hookrightarrow \hom(N,\prod_{i \in I} M_i)$ is mono, which identifies with $\bigoplus_i (M_i \otimes N) \hookrightarrow \bigoplus_{i \in I} \hom(N,M_i) \hookrightarrow \prod_{i \in I} \hom(N,M_i)$.

In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)

The next case would be to look at two cyclic modules, say $M = A/I$ and $N = A/J$ for two ideals $I,J$. Then $M \otimes N^* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$ and $\hom(N,M) \cong (I:J)/I$, and the canonical map is given by $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $\mathrm{Ann}(I) \to A/I$, $a \mapsto [a]$ with kernel $I \cap \mathrm{Ann}(I)$.

Here is how I would start.

Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that $M \otimes N^* \to \hom(N,M)$, $m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$ is a monomorphism. Both sides are additive functors in both variables. It follows formally:

1. The class $S$ is closed under finite direct sums (in both variables).
2. The class $S$ is closed under direct summands (in both variables).

Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).

We can also prove:

3. If $I$ is any index set and $(M_i,N) \in S$ for $i \in I$, then $(\bigoplus_{i \in I} M_i,N) \in S$:

This is because $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i)$ is mono iff the composition $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i) \hookrightarrow \hom(N,\prod_{i \in I} M_i)$ is mono, which identifies with $\bigoplus_i (M_i \otimes N) \hookrightarrow \bigoplus_{i \in I} \hom(N,M_i) \hookrightarrow \prod_{i \in I} \hom(N,M_i)$.

In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)

The next case would be to look at two cyclic modules, say $M = A/I$ and $N = A/J$ for two ideals $I,J$. Then $M \otimes N^* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$ and $\hom(N,M) \cong (I:J)/I$, and the canonical map is given by $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $\mathrm{Ann}(I) \to A/I$, $a \mapsto [a]$ with kernel $I \cap \mathrm{Ann}(I)$. This kernel can be non-trivial: take $A = k[X]/\langle X^2 \rangle$ and $I = \langle X \rangle$.

Here is how I would start.

Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that $M \otimes N^* \to \hom(N,M)$, $m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$ is a monomorphism. Both sides are additive functors in both variables. It follows formally:

1. The class $S$ is closed under finite direct sums (in both variables).
2. The class $S$ is closed under direct summands (in both variables).

Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).

We can also prove:

3. If $I$ is any index set and $(M_i,N) \in S$ for $i \in I$, then $(\bigoplus_{i \in I} M_i,N) \in S$:

This is because $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i)$ is mono iff the composition $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i) \hookrightarrow \hom(N,\prod_{i \in I} M_i)$ is mono, which identifies with $\bigoplus_i (M_i \otimes N) \hookrightarrow \bigoplus_{i \in I} \hom(N,M_i) \hookrightarrow \prod_{i \in I} \hom(N,M_i)$.

In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)

Here is how I would start.

Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that $M \otimes N^* \to \hom(N,M)$, $m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$ is a monomorphism. Both sides are additive functors in both variables. It follows formally:

1. The class $S$ is closed under finite direct sums (in both variables).
2. The class $S$ is closed under direct summands (in both variables).

Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).

We can also prove:

3. If $I$ is any index set and $(M_i,N) \in S$ for $i \in I$, then $(\bigoplus_{i \in I} M_i,N) \in S$:

This is because $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i)$ is mono iff the composition $(\bigoplus_{i \in I} M_i) \otimes N \to \hom(N,\bigoplus_{i \in I} M_i) \hookrightarrow \hom(N,\prod_{i \in I} M_i)$ is mono, which identifies with $\bigoplus_i (M_i \otimes N) \hookrightarrow \bigoplus_{i \in I} \hom(N,M_i) \hookrightarrow \prod_{i \in I} \hom(N,M_i)$.

In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)

The next case would be to look at two cyclic modules, say $M = A/I$ and $N = A/J$ for two ideals $I,J$. Then $M \otimes N^* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$ and $\hom(N,M) \cong (I:J)/I$, and the canonical map is given by $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $\mathrm{Ann}(I) \to A/I$, $a \mapsto [a]$ with kernel $I \cap \mathrm{Ann}(I)$.