I am following up on my comment.

The category of algebras over a commutative algebraic theory is monoidal closed. This is explained here http://ncatlab.org/nlab/show/commutative+algebraic+theory.

One can "commutativize" any theory by adding axioms expressing commutativity of any of its two operations (as in the nlab article). So, let us try to find a counterexample among categories of algebraic theories first, and then try to commutativize them.

Here is an example of a theory whose category of algebras may have non-injective coproduct injections: Take a theory which has $n$-ary terms $f, g, l$, $m$-ary terms $h$, $k$, and a binary term $p$, which satisfy

$$f(x_1, ..., x_n) = p(l(x_1, ..., x_n), h(y_1, ..., y_m))$$
$$g(x_1, ..., x_n) = p(l(x_1, ..., x_n), k(y_1, ..., y_m)).$$

Take a free algebra in this theory $F\{a_1, ..., a_n\}$ (shortly $F\{a\}$) on a set $\{a_1, ..., a_n\}$. Take a quotient of a free algebra $F\{b_1, ..., b_m\}/h(b_1, ..., b_m) \sim k(b_1, ..., b_m)$. In the coproduct of these two algebras we have

$$f(a) \sim p(l(a), h(b)) \sim p(l(a), k(b)) \sim g(a).$$

Thus, the coproduct injection from $F\{a\}$ is not injective (unless for some reason f(x) and g(x) equal to each other in the theory itself).

The theory of rings falls under this example by taking $f, g, l, h, k$ all the following zero terms $f = 2, g = 0, l = 1, h = 2, k = 0$, and $p(x, y) = xy$.

If we try to commutativize the theory of rings the constants get identified. In particular $2$ and $0$ get identified in the original theory, and we stay without a counterexample :).

In fact commutativization destroys the counterexample if the theory has any constant. Because, for a constant $c$, we should have $h(c, c, ...) = c = k(c, c, ...)$ by commutativity, and this forces $f(x)$ to equal $g(x)$ in the theory itself. This should be like this too, because the category of algebras of a commutative theory with a constant (necessarily unique) has a zero object, and thus has injective coproduct injections.

However, in other situations commutativization should be a harmless process. For example one can take a theory given by unary operation $f, g, l, h, k$ and a binary operation $p$, with the only axioms the above given two equations. Commutativazation of this theory should give a counterexample to the question.

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