It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to $$[X,Y]=uX +vY +cI$$ then the desired logarithm of the product of matrix exponentials equals $$\log(e^X e^{tY})=tY+X+f(u,v,t)(uX +vY +cI)$$ $$f(u,v,t)=\frac{(ut-v)e^{ut+v}-ute^{ut}+ve^{v}}{uv(e^{ut}-e^{v})}$$ The large-$t$ limit can now be read off once the sign of $u$ is known: $$\lim_{t\rightarrow\infty}f(u,v,t)=\begin{cases} -1/u&\text{if}\;u<0\\ (t/v)(e^v-1)&\text{if}\;u>0\\ (t/v)[v-1+v(e^v-1)^{-1}]&\text{if}\;u=0 \end{cases}$$ There are no terms greater than order $t$ in the large-$t$ limit for this class of commutators.

It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to $$[X,Y]=uX +vY +cI$$ then the desired logarithm of the product of matrix exponentials equals $$\log(e^X e^{tY})=tY+X+f(u,v,t)(uX +vY +cI)$$ $$f(u,v,t)=\frac{(ut-v)e^{ut+v}-ute^{ut}+ve^{v}}{uv(e^{ut}-e^{v})}$$ The large-$t$ limit can now be read off once the sign of $u$ is known: $$\lim_{t\rightarrow\infty}f(u,v,t)=\begin{cases} -1/u&\text{if}\;u<0\\ (t/v)(e^v-1)&\text{if}\;u>0\\ (t/v)[v-1+v(e^v-1)^{-1}]&\text{if}\;u=0 \end{cases}$$ There are no terms greater than order $t$ in the large-$t$ limit for this class of commutators. (I'm actually a bit puzzled how $t^2$ terms and higher might appear at all, an explicit example would help me a lot.)

It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to $$[X,Y]=uX +vY +cI$$ then the desired logarithm of the product of matrix exponentials equals $$\log(e^X e^{tY})=tY+X+f(u,v,t)(uX +vY +cI)$$ $$f(u,v,t)=\frac{(ut-v)e^{ut+v}-ute^{ut}+ve^{v}}{uv(e^{ut}-e^{v})}$$ The large-$t$ limit can now be read off once the sign of $u$ is known: $$\lim_{t\rightarrow\infty}f(u,v,t)=\begin{cases} -1/u&\text{if}\;u<0\\ (t/v)(e^v-1)&\text{if}\;u>0\\ (t/v)[v-1+v(e^v-1)^{-1}]&\text{if}\;u=0 \end{cases}$$

It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to $$[X,Y]=uX +vY +cI$$ then the desired logarithm of the product of matrix exponentials equals $$\log(e^X e^{tY})=tY+X+f(u,v,t)(uX +vY +cI)$$ $$f(u,v,t)=\frac{(ut-v)e^{ut+v}-ute^{ut}+ve^{v}}{uv(e^{ut}-e^{v})}$$ The large-$t$ limit can now be read off once the sign of $u$ is known: $$\lim_{t\rightarrow\infty}f(u,v,t)=\begin{cases} -1/u&\text{if}\;u<0\\ (t/v)(e^v-1)&\text{if}\;u>0\\ (t/v)[v-1+v(e^v-1)^{-1}]&\text{if}\;u=0 \end{cases}$$ There are no terms greater than order $t$ in the large-$t$ limit for this class of commutators.