Convolution of shifted signal
From your confusion of $x(t+a) \star h(t+a)$ vs $h(t) \star x(t+a)$ 
I guess that a little help on the argument manipulations on functions and convolutions could be appropriate here, moving in simple examples:

First let us express the usual simplistic case.  Consider the relation: 
$$ y(t) = h(t) \cdot x(t) + g(t) \tag{1}$$

then manipulations on the argument $t$ is applied to all functions on both sides

$$ y(t+a) = h(t+a) \cdot x(t+a) + g(t+a)$$

or an arbitrary transform on $t$ would similary be:
$$ y(\phi(t)) = h(\phi(t)) \cdot x(\phi(t)) + g(\phi(t))$$

Now consider the case where two functions convolved to produce the third:
$$y(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau) d\tau $$

which is abbreviated as
$$ y(t) = h(t) \star x(t) \tag{2} $$

Now be careful to interpret the case-2. The variable $t$ shows in all functions as an argument but you may not apply the transform on $t$ as you did with the case-1, so assume you have an arbitrary transform on $t$ as $\phi(t)$ then

$$ y(\phi(t)) \neq h(\phi(t)) \star x(\phi(t)) \tag{3} $$

For example, as in your case, if $\phi(t) = t+a$ then you get 

$$ y(t+a) \neq h(t+a) \star x(t+a)$$
but
$$ y(t+a) = h(t) \star x(t+a) = h(t+a) \star x(t) $$

The justification of this can (only) be seen when you consider the integral definition of the convolution operator:

$$
\begin{align}
y(t) &amp;= h(t) \star x(t) \\
&amp; = \int_{-\infty}^{\infty} h(\tau) x(t-\tau)d\tau \\
y(t+a) &amp; = \int_{-\infty}^{\infty} h(\tau) x((t+a)-\tau)d\tau \\
&amp; = h(t) \star x(t+a) \\
\end{align}
$$

Note that since the live variable inside the integral only happens in one function ($x(t-\tau)$ in this case) then a change in $t$ will only affect one of them and you get:

$$ y(t+a) = h(t) \star x(t+a) $$
or from commutativity of convolution you get
$$ y(t+a) = h(t+a) \star x(t) $$

So this provides the answer you were looking for. However, it's not over. Because the following case represents an exception:

$$ y(-t) \neq h(t) \star x(-t) $$
but
$$ y(-t) = h(-t) \star x(-t) \tag{4} $$

So how to see this case-4. Again, using the integral definition :

Assuming that $y(t) = h(t) \star x(t)$, then compute the convolution between two new signals $g(t)=h(-t)$ and $z(t)=x(-t)$ as:
$$
\begin{align}
w(t) &amp;= g(t) \star z(t) \\
&amp; = \int_{-\infty}^{\infty} g(\tau) z(t-\tau)d\tau &amp;g(\tau)=h(-\tau),z(t-\tau)=x(-(t-\tau)) \\
&amp; = \int_{-\infty}^{\infty} h(-\tau) x(-(t-\tau))d\tau  &amp;\text{ let } \tau'=-\tau \\
&amp; = -\int_{\infty}^{-\infty} h(\tau') x(-(t+\tau'))d\tau'  &amp;\text{ replace } \tau'  \text{ with } \tau \\
&amp; = \int_{-\infty}^{\infty} h(\tau) x(-t-\tau) d\tau  \\
&amp; = y(-t) \\
\end{align}
$$

hence we conclude that $h(-t) \star x(-t) = w(t) = y(-t) $. As stated before, you must always consult to the (explicit) integral definition to decide on the correct functions used in the convolution operator.